Quick Answer
SAT parabola questions test how well you can connect vertex form, standard form, graphs, axis of symmetry, roots, and real-world models. This page includes 65 SAT-style parabola questions with answer choices, full solutions, and trap notes. Beginning with simple vertex and intercept identification, the problems progress to more difficult square completion, transformation, and parameter issues.
What to Know Before You Start
- The graph of a quadratic function is called a parabola. Depending on the leading coefficient's sign, its U-shaped opening can be either upward or downward.
- The most common SAT forms are standard form f(x) = ax² + bx + c and vertex form f(x) = a(x − h)² + k, where (h, k) is the vertex.
- The vertex is the minimum point when a > 0 and the maximum point when a < 0.
- The axis of symmetry passes through the vertex and has equation x = h or x = −b/(2a).
- The points on the x-axis where the parabola crosses are known as the x-intercepts (roots, zeros). To find these, solve with y = 0.
- The y-intercept is the value of c in standard form, found by setting x = 0.
- Many incorrect responses result from mistaking the direction of opening, confusing h with −h in vertex form, or forgetting that the vertex formula provides an x-value that needs to be substituted back to determine y.
In This Guide – 65 SAT Parabola Practice Questions
- What does the SAT test in parabolas?
- How do vertex form questions work?
- How do you handle standard form and intercepts?
- How are parabolas used in SAT word problems?
- How do you handle transformations and equivalent forms?
- What do hard SAT parabola questions look like?
- What mistakes cost students points on parabolas?
- How should you study SAT parabolas in 2 weeks?
- Frequently asked questions
Start With SAT Math Topic-Wise Practice
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Use the SAT Prep Guide E-Book to plan SAT Math practice, review common parabola traps, and connect every practice set with a weekly score improvement plan.
Download SAT Prep Guide E-BookWhat Does the SAT Test in Parabolas?
The SAT Math Advanced Math domain includes parabolas. They could show up on the Digital SAT as vertex form equations, standard form equations, projectile motion models, graphs with labeled features, or comparable expressions that call for square completion. Although the math is typically short, the question phrasing frequently requires you to determine the true meaning of the vertices, roots, and coefficients.
Prior to solution, a proficient SAT student undertakes three tasks: determines the equation's form, finds the necessary vertex or intercepts, and links any coefficients to their geometric meaning. Most parabola problems become simple algebraic problems if those components are understood.
| SAT Parabola Skill | What It Tests | Common Trap | Practice Set |
| Vertex form basics | Reading vertex, direction, and axis of symmetry from f(x) = a(x − h)² + k | Confusing the sign of h | Q1–Q15 |
| Standard form and intercepts | Finding vertex from −b/(2a), y-intercept, and roots | Forgetting to find the y-coordinate of the vertex | Q16–Q30 |
| Parabola word problems | Projectile motion, area optimization, revenue models | Using the wrong variable for the answer | Q31–Q45 |
| Transformations and equivalent forms | Converting between forms, completing the square, graph shifts | Sign errors when completing the square | Q46–Q55 |
| Hard mixed questions | Parameters, system intersections, discriminant, multi-step models | Not connecting the discriminant to the number of solutions | Q56–Q65 |
SAT strategy: Find out whether the equation is in standard form or vertex form before using the calculator. Identify whether the question is asking for a vertex, an intercept, the meaning of a coefficient, or an equivalent expression.
How Do SAT Parabola Questions Test Vertex Form?
To begin, use the vertex form f(x) = a(x − h)² + k. These questions test your ability to determine the axis of symmetry x = h, recognize the vertex (h, k), and deduce the direction of opening from the sign of a.
Vertex Form Basics · Easy · Question 1
The function f(x) = 2(x − 3)² + 5 represents a parabola. What are the coordinates of the vertex?
A) (−3, 5) B) (3, 5) C) (3, −5) D) (5, 3)
Show full solution
Solution: In vertex form f(x) = a(x − h)² + k, the vertex is (h, k).
Here, f(x) = 2(x − 3)² + 5, so h = 3 and k = 5.
The vertex is (3, 5).
Trap note: Choice A uses −3 instead of 3. Remember that (x − 3) means h = +3, not −3.
Answer: B
Vertex Form Basics · Easy · Question 2
The parabola g(x) = −(x + 4)² + 9 opens in which direction?
A) Upward B) Downward C) To the right D) To the left
Show full solution
Solution: The coefficient a in front of the squared term determines the direction.
Here, a = −1 (negative), so the parabola opens downward.
Trap note: Choices C and D describe horizontal parabolas, which are not functions. SAT parabolas always open up or down.
Answer: B
Vertex Form Basics · Easy · Question 3
What is the equation of the axis of symmetry for f(x) = 3(x − 7)² − 2?
A) x = −7 B) x = 7 C) x = −2 D) y = −2
Show full solution
Solution: The axis of symmetry passes through the vertex and is a vertical line x = h.
From f(x) = 3(x − 7)² − 2, we have h = 7.
The axis of symmetry is x = 7.
Trap note: Choice D gives y = −2, which is a horizontal line—not an axis of symmetry for a vertical parabola.
Answer: B
Vertex Form Basics · Easy · Question 4
A parabola has vertex (−2, 6) and passes through the point (0, 14). Which equation represents the parabola?
A) f(x) = 2(x + 2)² + 6 B) f(x) = 2(x − 2)² + 6 C) f(x) = (x + 2)² + 6 D) f(x) = 4(x + 2)² + 6
Show full solution
Solution: Start with vertex form: f(x) = a(x − h)² + k = a(x − (−2))² + 6 = a(x + 2)² + 6
Use the point (0, 14): 14 = a(0 + 2)² + 6
14 = 4a + 6 → 8 = 4a → a = 2
The equation is f(x) = 2(x + 2)² + 6.
Answer: A
Vertex Form Basics · Medium · Question 5
For the function f(x) = −2(x − 1)² + 8, what is the maximum value of f(x)?
A) −2 B) 1 C) 6 D) 8
Show full solution
Solution: Since a = −2 < 0, the parabola opens downward, so the vertex is a maximum.
The vertex is at (1, 8), so the maximum value of f(x) is 8.
Trap note: Choice B gives the x-coordinate of the vertex, not the maximum y-value.
Answer: D
Vertex Form Basics · Easy · Question 6
If f(x) = (x − 5)² − 16, what is the value of f(5)?
A) −16 B) −11 C) 0 D) 9
Show full solution
Solution: f(5) = (5 − 5)² − 16 = 0² − 16 = −16
Note: x = 5 is the vertex, so f(5) equals the k-value.
Answer: A
Vertex Form Basics · Medium · Question 7
The graph of f(x) = (x + 3)² is shifted 4 units up. Which equation represents the new function?
A) f(x) = (x + 3)² + 4 B) f(x) = (x + 3)² − 4 C) f(x) = (x + 7)² D) f(x) = (x − 1)²
Show full solution
Solution: A vertical shift of 4 units up adds 4 to the function.
New function: f(x) = (x + 3)² + 4
Trap note: Choices C and D show horizontal shifts, not vertical shifts.
Answer: A
Vertex Form Basics · Medium · Question 8
For what value of x does f(x) = 4(x − 2)² − 7 reach its minimum?
A) −7 B) −2 C) 2 D) 4
Show full solution
Solution: The minimum occurs at the vertex. In vertex form f(x) = a(x − h)² + k, the x-coordinate of the vertex is h.
Here, h = 2, so the minimum occurs at x = 2.
Trap note: Choice A gives the minimum value, not the x-value where the minimum occurs.
Answer: C
Vertex Form Basics · Medium · Question 9
A parabola has vertex (4, −3) and opens upward. Which could be its equation?
A) f(x) = −(x − 4)² − 3 B) f(x) = (x − 4)² − 3 C) f(x) = (x + 4)² − 3 D) f(x) = (x − 4)² + 3
Show full solution
Solution: Vertex (4, −3) means h = 4 and k = −3.
Opens upward means a > 0.
The equation is f(x) = (x − 4)² − 3 (or any positive multiple).
Trap note: Choice A opens downward (a < 0). Choice C has wrong vertex (−4, −3).
Answer: B
Vertex Form Basics · Hard · Question 10
If f(x) = a(x − 3)² + 5 and the graph passes through (5, 13), what is the y-intercept of f?
A) 5 B) 13 C) 23 D) 32
Show full solution
Solution: First find a using point (5, 13):
13 = a(5 − 3)² + 5 → 13 = 4a + 5 → a = 2
So f(x) = 2(x − 3)² + 5
Find y-intercept by setting x = 0:
f(0) = 2(0 − 3)² + 5 = 2(9) + 5 = 18 + 5 = 23
Answer: C
Vertex Form Basics · Easy · Question 11
Which parabola has its vertex in the third quadrant?
A) f(x) = (x + 2)² + 3 B) f(x) = (x − 2)² − 3 C) f(x) = (x + 2)² − 3 D) f(x) = (x − 2)² + 3
Show full solution
Solution: Third quadrant means x < 0 and y < 0.
Check each vertex:
A: (−2, 3) — Quadrant II
B: (2, −3) — Quadrant IV
C: (−2, −3) — Quadrant III ✓
D: (2, 3) — Quadrant I
Answer: C
Vertex Form Basics · Medium · Question 12
The function g(x) = 3(x − 1)² − 12 has how many x-intercepts?
A) 0 B) 1 C) 2 D) 3
Show full solution
Solution: Set g(x) = 0:
3(x − 1)² − 12 = 0 → 3(x − 1)² = 12 → (x − 1)² = 4
x − 1 = ±2 → x = 3 or x = −1
The parabola has 2 x-intercepts.
Answer: C
Vertex Form Basics · Medium · Question 13
If the vertex of f(x) = a(x − h)² + k is (−5, 2) and f(−3) = 10, what is the value of a?
A) 1 B) 2 C) 3 D) 4
Show full solution
Solution: f(x) = a(x + 5)² + 2
Use f(−3) = 10:
10 = a(−3 + 5)² + 2 → 10 = 4a + 2 → 8 = 4a → a = 2
Answer: B
Vertex Form Basics · Hard · Question 14
The range of f(x) = −2(x + 1)² + 10 is:
A) y ≤ 10 B) y ≥ 10 C) y ≤ −1 D) All real numbers
Show full solution
Solution: Since a = −2 < 0, the parabola opens downward.
The vertex (−1, 10) is the maximum point.
The range is y ≤ 10 (all y-values at or below the maximum).
Trap note: Choice B would be correct if the parabola opened upward.
Answer: A
Vertex Form Basics · Hard · Question 15
If f(x) = (x − 4)² − 9, what are the x-intercepts?
A) x = 1 and x = 7 B) x = −1 and x = 7 C) x = 1 and x = −7 D) x = 4 only
Show full solution
Solution: Set f(x) = 0:
(x − 4)² − 9 = 0 → (x − 4)² = 9 → x − 4 = ±3
x = 4 + 3 = 7 or x = 4 − 3 = 1
The x-intercepts are x = 1 and x = 7.
Answer: A
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How Do You Handle Standard Form and Intercepts on the SAT?
The usual form f(x) = ax² + bx + c is the subject of these questions. You must determine the y-intercept (the c value), locate the vertex using x = −b/(2a), then find roots by factoring or applying the quadratic formula.
Standard Form & Intercepts · Easy · Question 16
What is the y-intercept of f(x) = x² − 6x + 8?
A) −6 B) 1 C) 8 D) −8
Show full solution
Solution: The y-intercept is the value when x = 0, which is the constant term c.
In f(x) = x² − 6x + 8, c = 8.
Answer: C
Standard Form & Intercepts · Easy · Question 17
What is the x-coordinate of the vertex of f(x) = x² + 8x + 12?
A) −8 B) −4 C) 4 D) 8
Show full solution
Solution: Use x = −b/(2a).
Here a = 1 and b = 8, so x = −8/(2·1) = −8/2 = −4
Trap note: Choice A forgets to divide by 2a.
Answer: B
Standard Form & Intercepts · Medium · Question 18
What are the x-intercepts of f(x) = x² − 5x + 6?
A) x = 1 and x = 6 B) x = 2 and x = 3 C) x = −2 and x = −3 D) x = 5 only
Show full solution
Solution: Factor: x² − 5x + 6 = (x − 2)(x − 3) = 0
x = 2 or x = 3
The x-intercepts are x = 2 and x = 3.
Trap note: Choice C has wrong signs—the factors are (x − 2) and (x − 3), not (x + 2) and (x + 3).
Answer: B
Standard Form & Intercepts · Medium · Question 19
What is the vertex of f(x) = x² − 4x + 7?
A) (2, 3) B) (2, 7) C) (−2, 15) D) (4, 7)
Show full solution
Solution: x-coordinate: x = −(−4)/(2·1) = 4/2 = 2
y-coordinate: f(2) = (2)² − 4(2) + 7 = 4 − 8 + 7 = 3
Vertex is (2, 3).
Trap note: Choice B uses c = 7 as the y-coordinate without substituting.
Answer: A
Standard Form & Intercepts · Medium · Question 20
For f(x) = 2x² + 12x + 10, what is the axis of symmetry?
A) x = −6 B) x = −3 C) x = 3 D) x = 6
Show full solution
Solution: Axis of symmetry: x = −b/(2a) = −12/(2·2) = −12/4 = −3
Trap note: Choice A forgets to divide by 2a.
Answer: B
Standard Form & Intercepts · Easy · Question 21
If f(x) = x² + 4x + 4, which is true?
A) f has two x-intercepts B) f has one x-intercept C) f has no x-intercepts D) f opens downward
Show full solution
Solution: Factor: x² + 4x + 4 = (x + 2)² = 0
x = −2 (double root)
The parabola touches the x-axis at exactly one point.
Answer: B
Standard Form & Intercepts · Medium · Question 22
The discriminant of f(x) = 3x² − 2x + 5 is:
A) −56 B) −4 C) 4 D) 64
Show full solution
Solution: Discriminant = b² − 4ac = (−2)² − 4(3)(5) = 4 − 60 = −56
Since the discriminant is negative, this parabola has no real x-intercepts.
Answer: A
Standard Form & Intercepts · Hard · Question 23
If f(x) = −x² + 6x − 5, what is the maximum value of f(x)?
A) 3 B) 4 C) 5 D) 6
Show full solution
Solution: Since a = −1 < 0, the parabola opens downward, so the vertex gives the maximum.
x = −b/(2a) = −6/(2·(−1)) = −6/(−2) = 3
f(3) = −(3)² + 6(3) − 5 = −9 + 18 − 5 = 4
Trap note: Choice A is the x-value, not the maximum y-value.
Answer: B
Standard Form & Intercepts · Medium · Question 24
If a parabola has x-intercepts at x = −1 and x = 5, what is the x-coordinate of the vertex?
A) 2 B) 3 C) 4 D) 6
Show full solution
Solution: The vertex lies on the axis of symmetry, which is halfway between the x-intercepts.
x = (−1 + 5)/2 = 4/2 = 2
Answer: A
Standard Form & Intercepts · Medium · Question 25
Which equation represents a parabola with x-intercepts at 2 and 6?
A) f(x) = (x + 2)(x + 6) B) f(x) = (x − 2)(x − 6) C) f(x) = (x − 2)(x + 6) D) f(x) = x² − 8
Show full solution
Solution: If x-intercepts are 2 and 6, then the factored form is f(x) = a(x − 2)(x − 6).
With a = 1: f(x) = (x − 2)(x − 6)
Trap note: Choice A has intercepts at −2 and −6 (wrong signs).
Answer: B
Standard Form & Intercepts · Hard · Question 26
For what value of c does f(x) = x² − 8x + c have exactly one x-intercept?
A) 8 B) 16 C) 32 D) 64
Show full solution
Solution: Exactly one x-intercept means discriminant = 0.
b² − 4ac = 0 → (−8)² − 4(1)(c) = 0 → 64 − 4c = 0 → c = 16
Answer: B
Standard Form & Intercepts · Easy · Question 27
If f(x) = 2x² − 8x + 6, what is f(0)?
A) −8 B) 0 C) 2 D) 6
Show full solution
Solution: f(0) = 2(0)² − 8(0) + 6 = 6
This is the y-intercept, which equals c in standard form.
Answer: D
Standard Form & Intercepts · Medium · Question 28
The sum of the x-intercepts of f(x) = x² − 7x + 10 is:
A) 2 B) 5 C) 7 D) 10
Show full solution
Solution: By Vieta’s formulas, for ax² + bx + c = 0, sum of roots = −b/a.
Sum = −(−7)/1 = 7
Or factor: (x − 2)(x − 5) = 0, roots are 2 and 5, sum = 7.
Answer: C
Standard Form & Intercepts · Hard · Question 29
If f(x) = ax² + bx + c has vertex (3, −4) and passes through (0, 5), what is the value of a?
A) 1 B) 2 C) 3 D) 4
Show full solution
Solution: Use vertex form: f(x) = a(x − 3)² − 4
Point (0, 5): 5 = a(0 − 3)² − 4 → 5 = 9a − 4 → 9 = 9a → a = 1
Answer: A
Standard Form & Intercepts · Medium · Question 30
The product of the x-intercepts of f(x) = x² + 2x − 15 is:
A) −15 B) −2 C) 2 D) 15
Show full solution
Solution: By Vieta’s formulas, product of roots = c/a = −15/1 = −15
Or factor: (x + 5)(x − 3) = 0, roots are −5 and 3, product = −15.
Answer: A
How Are Parabolas Used in SAT Word Problems?
Parabolas are used in SAT word problems to simulate revenue/profit models, area optimization (maximizing area), and projectile motion (height vs. time). Determining whether you require the vertices (maximum/minimum), the roots (when something hits the ground or breaks even), or a certain function value is crucial.
Parabola Word Problems · Easy · Question 31
A ball is hurled upward. The formula for its height h in feet after t seconds is h(t) = −16t² + 64t + 5. What was the ball's starting height?
A) 5 feet B) 16 feet C) 64 feet D) 69 feet
Show full solution
Solution: Initial height is when t = 0.
h(0) = −16(0)² + 64(0) + 5 = 5 feet
Trap note: Choice C is the coefficient of t (related to initial velocity), not the initial height.
Answer: A
Parabola Word Problems · Medium · Question 32
Using h(t) = −16t² + 64t + 5 from Question 31, what is the maximum height reached by the ball?
A) 64 feet B) 69 feet C) 85 feet D) 128 feet
Show full solution
Solution: Maximum height occurs at the vertex. Since a < 0, the parabola opens downward.
t = −b/(2a) = −64/(2·(−16)) = −64/(−32) = 2 seconds
h(2) = −16(4) + 64(2) + 5 = −64 + 128 + 5 = 69 feet
Answer: B
Parabola Word Problems · Medium · Question 33
At what time does the ball in Question 31 reach its maximum height?
A) 1 second B) 2 seconds C) 4 seconds D) 5 seconds
Show full solution
Solution: Time at maximum is the x-coordinate of the vertex.
t = −b/(2a) = −64/(2·(−16)) = 2 seconds
Trap note: Don’t confuse the time to reach maximum (2 sec) with the maximum height (69 feet).
Answer: B
Parabola Word Problems · Easy · Question 34
The formula for a company's profit P in dollars is P(x) = −2x² + 80x − 600, where x is the quantity of goods sold. To maximize profit, how many things must be sold
A) 10 B) 20 C) 40 D) 80
Show full solution
Solution: Maximum profit at the vertex.
x = −b/(2a) = −80/(2·(−2)) = −80/(−4) = 20 items
Answer: B
Parabola Word Problems · Medium · Question 35
Using P(x) = −2x² + 80x − 600 from Question 34, what is the maximum profit?
A) $200 B) $400 C) $600 D) $800
Show full solution
Solution: P(20) = −2(20)² + 80(20) − 600
= −2(400) + 1600 − 600 = −800 + 1600 − 600 = $200
Answer: A
Parabola Word Problems · Hard · Question 36
The perimeter of a rectangular garden is forty meters. A(x) = x(20 − x) gives the area A if one side is x meters long. Which measurements make the most space?
A) 5 m × 15 m B) 8 m × 12 m C) 10 m × 10 m D) 4 m × 16 m
Show full solution
Solution: A(x) = x(20 − x) = −x² + 20x
Maximum at x = −20/(2·(−1)) = 10
Other side = 20 − 10 = 10
Dimensions: 10 m × 10 m (a square maximizes area for a given perimeter)
Answer: C
Parabola Word Problems · Medium · Question 37
A rocket's height h in meters after t seconds is h(t) = −5t² + 30t. When does the rocket hit the ground?
A) 3 seconds B) 5 seconds C) 6 seconds D) 30 seconds
Show full solution
Solution: Set h(t) = 0: −5t² + 30t = 0
−5t(t − 6) = 0 → t = 0 or t = 6
t = 0 is launch, so the rocket hits the ground at t = 6 seconds.
Answer: C
Parabola Word Problems · Medium · Question 38
Using h(t) = −5t² + 30t from Question 37, what is the maximum height of the rocket?
A) 30 meters B) 45 meters C) 90 meters D) 180 meters
Show full solution
Solution: t at max = −30/(2·(−5)) = 3 seconds
h(3) = −5(9) + 30(3) = −45 + 90 = 45 meters
Answer: B
Parabola Word Problems · Hard · Question 39
A store finds that if it charges p dollars per item, it sells (100 − 2p) items per day. Revenue R = p(100 − 2p). What price maximizes revenue?
A) $20 B) $25 C) $40 D) $50
Show full solution
Solution: R = p(100 − 2p) = −2p² + 100p
p = −100/(2·(−2)) = 100/4 = $25
Answer: B
Parabola Word Problems · Easy · Question 40
An object is dropped from a height of 100 feet. Its height after t seconds is h(t) = −16t² + 100. After how many seconds does it hit the ground?
A) 2 seconds B) 2.5 seconds C) 3 seconds D) 6.25 seconds
Show full solution
Solution: Set h(t) = 0: −16t² + 100 = 0
16t² = 100 → t² = 6.25 → t = 2.5 seconds
Answer: B
Parabola Word Problems · Medium · Question 41
The height of a fountain's water spray is modeled by h(x) = −0.5x² + 4x, where x is the horizontal distance from the fountain. What is the maximum height of the water?
A) 4 feet B) 8 feet C) 16 feet D) 32 feet
Show full solution
Solution: x = −4/(2·(−0.5)) = 4
h(4) = −0.5(16) + 4(4) = −8 + 16 = 8 feet
Answer: B
Parabola Word Problems · Hard · Question 42
A company's cost C and revenue R are given by C(x) = 2x + 100 and R(x) = −x² + 50x. At what value of x is profit (R − C) maximized?
A) 24 B) 25 C) 26 D) 48
Show full solution
Solution: Profit P(x) = R(x) − C(x) = (−x² + 50x) − (2x + 100) = −x² + 48x − 100
x = −48/(2·(−1)) = 24
Answer: A
Parabola Word Problems · Easy · Question 43
A ball's height is h(t) = −4.9t² + 19.6t + 1. What does the coefficient −4.9 represent?
A) Initial velocity B) Half the acceleration due to gravity C) Maximum height D) Time to reach maximum
Show full solution
Solution: In projectile motion h(t) = −(1/2)gt² + v₀t + h₀, the coefficient of t² is −(1/2)g.
Here, −4.9 = −(1/2)(9.8), representing half the acceleration due to gravity (in meters).
Answer: B
Parabola Word Problems · Medium · Question 44
Two balls are thrown. Ball A: h(t) = −16t² + 48t. Ball B: h(t) = −16t² + 64t. Which ball reaches a greater maximum height and by how much?
A) Ball A by 16 feet B) Ball A by 28 feet C) Ball B by 28 feet D) Ball B by 36 feet
Show full solution
Solution:
Ball A: t = 48/32 = 1.5, h(1.5) = −16(2.25) + 48(1.5) = −36 + 72 = 36 feet
Ball B: t = 64/32 = 2, h(2) = −16(4) + 64(2) = −64 + 128 = 64 feet
Ball B by 64 − 36 = 28 feet
Answer: C
Parabola Word Problems · Hard · Question 45
A wire 60 cm long is bent into a rectangle. What is the maximum area that can be enclosed?
A) 200 cm² B) 225 cm² C) 300 cm² D) 400 cm²
Show full solution
Solution: Let width = x, then length = (60 − 2x)/2 = 30 − x
A = x(30 − x) = −x² + 30x
Max at x = 15, A = 15(15) = 225 cm²
Answer: B
How Do You Handle Transformations and Equivalent Forms?
The same parabola is frequently given in several formats by the SAT. You might have to compare equivalent equations, find graph shifts, complete the square, or convert between forms.
Transformations & Forms · Medium · Question 46
Which expression is equivalent to x² + 6x + 5?
A) (x + 3)² − 4 B) (x + 3)² + 4 C) (x + 5)(x + 1) D) Both A and C
Show full solution
Solution:
Check A: (x + 3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5 ✓
Check C: (x + 5)(x + 1) = x² + 6x + 5 ✓
Both A and C are equivalent.
Answer: D
Transformations & Forms · Medium · Question 47
Write x² − 10x + 21 in vertex form.
A) (x − 5)² − 4 B) (x − 5)² + 4 C) (x + 5)² − 4 D) (x − 10)² + 21
Show full solution
Solution: Complete the square:
x² − 10x + 21 = (x² − 10x + 25) − 25 + 21 = (x − 5)² − 4
Answer: A
Transformations & Forms · Easy · Question 48
The graph of g(x) = x² is shifted 3 units right. Which equation represents the new graph?
A) g(x) = (x + 3)² B) g(x) = (x − 3)² C) g(x) = x² + 3 D) g(x) = x² − 3
Show full solution
Solution: A horizontal shift right by h units replaces x with (x − h).
Shift right 3: g(x) = (x − 3)²
Trap note: Choices C and D are vertical shifts, not horizontal.
Answer: B
Transformations & Forms · Medium · Question 49
If f(x) = x², which transformation gives g(x) = 2x² − 3?
A) Vertical stretch by 2, shift down 3 B) Horizontal stretch by 2, shift down 3 C) Vertical stretch by 2, shift up 3 D) Vertical compression by 2, shift down 3
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Solution: g(x) = 2x² − 3 = 2·f(x) − 3
The 2 multiplies the output (vertical stretch by factor 2).
The −3 subtracts from the output (shift down 3).
Answer: A
Transformations & Forms · Hard · Question 50
Write 2x² + 8x + 5 in vertex form.
A) 2(x + 2)² − 3 B) 2(x + 2)² + 3 C) 2(x − 2)² − 3 D) (x + 4)² + 5
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Solution: Factor out 2 from the first two terms:
2x² + 8x + 5 = 2(x² + 4x) + 5
Complete the square inside: 2(x² + 4x + 4 − 4) + 5 = 2((x + 2)² − 4) + 5
= 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3
Answer: A
Transformations & Forms · Easy · Question 51
The graph of f(x) = x² is reflected across the x-axis. Which equation represents the new graph?
A) f(x) = x² B) f(x) = −x² C) f(x) = (−x)² D) f(x) = |x²|
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Solution: Reflection across the x-axis negates the y-values.
f(x) = −x²
Trap note: (−x)² = x², so choice C is just the original function.
Answer: B
Transformations & Forms · Medium · Question 52
Which equation is equivalent to f(x) = (x − 4)(x + 2)?
A) f(x) = x² − 2x − 8 B) f(x) = x² + 2x − 8 C) f(x) = x² − 6x − 8 D) f(x) = (x − 1)² − 9
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Solution: (x − 4)(x + 2) = x² + 2x − 4x − 8 = x² − 2x − 8
Check D: (x − 1)² − 9 = x² − 2x + 1 − 9 = x² − 2x − 8 ✓
Both A and D are correct (but only A is shown as a choice).
Answer: A
Transformations & Forms · Hard · Question 53
The function f(x) = a(x − h)² + k has vertex (2, −5) and passes through (4, 3). What is the value of a?
A) 1 B) 2 C) 3 D) 4
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Solution: f(x) = a(x − 2)² − 5
Use (4, 3): 3 = a(4 − 2)² − 5 → 3 = 4a − 5 → 8 = 4a → a = 2
Answer: B
Transformations & Forms · Medium · Question 54
If f(x) = x² and g(x) = f(x − 2) + 3, what are the coordinates of the vertex of g?
A) (−2, 3) B) (2, 3) C) (2, −3) D) (3, 2)
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Solution: g(x) = (x − 2)² + 3
This is vertex form with h = 2 and k = 3.
Vertex: (2, 3)
Answer: B
Transformations & Forms · Hard · Question 55
For what value of k does x² + kx + 9 have exactly one x-intercept?
A) k = 3 only B) k = 6 only C) k = ±6 D) k = ±3
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Solution: Exactly one x-intercept means discriminant = 0.
k² − 4(1)(9) = 0 → k² = 36 → k = ±6
Answer: C
What Do Hard SAT Parabola Questions Look Like?
Harder questions combine function notation, parameter reasoning, system intersections, discriminant analysis, and real-world interpretation. The math is still quadratic, but the setup is less direct.
Hard Mixed Questions · Hard · Question 56
The line y = 2x + k is tangent to the parabola y = x². What is the value of k?
A) −1 B) 0 C) 1 D) 2
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Solution: Tangent means exactly one intersection point.
Set equal: x² = 2x + k → x² − 2x − k = 0
One solution means discriminant = 0: 4 − 4(1)(−k) = 0 → 4 + 4k = 0 → k = −1
Answer: A
Hard Mixed Questions · Hard · Question 57
If f(x) = x² − 4x + 3 and g(x) = x − 1, at what points do the graphs intersect?
A) (1, 0) and (4, 3) B) (0, 3) and (3, 2) C) (1, 0) and (3, 0) D) (2, 1) only
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Solution: Set f(x) = g(x): x² − 4x + 3 = x − 1
x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0 → x = 1 or x = 4
At x = 1: y = 1 − 1 = 0 → (1, 0)
At x = 4: y = 4 − 1 = 3 → (4, 3)
Intersection points: (1, 0) and (4, 3)
Answer: A
Hard Mixed Questions · Hard · Question 58
For what values of m does the system y = x² and y = mx − 4 have no real solutions?
A) −4 < m < 4 B) m < −4 or m > 4 C) m = 0 D) All values of m
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Solution: Set the equations equal: x² = mx − 4, so x² − mx + 4 = 0.
For no real solutions, the discriminant must be less than 0: m² − 16 < 0.
Therefore, −4 < m < 4.
Answer: A
Hard Mixed Questions · Hard · Question 59
If f(x) = x² − 6x + c has a minimum value of −2, what is c?
A) 5 B) 7 C) 9 D) 11
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Solution: Vertex x-coordinate: x = 6/2 = 3
Minimum value f(3) = 9 − 18 + c = c − 9 = −2
c = 7
Answer: B
Hard Mixed Questions · Hard · Question 60
The parabola y = ax² + bx + c has vertex (−3, 4) and passes through the origin. What is the value of a?
A) −4/9 B) −1/9 C) 4/9 D) 1/9
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Solution: Vertex form: y = a(x + 3)² + 4
Passes through (0, 0): 0 = a(9) + 4 → 9a = −4 → a = −4/9
Answer: A
Hard Mixed Questions · Hard · Question 61
If f(x) = x² and g(x) = f(2x), how does the graph of g compare to f?
A) Horizontally stretched by factor 2 B) Horizontally compressed by factor 2 C) Vertically stretched by factor 4 D) Both B and C
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Solution: g(x) = f(2x) = (2x)² = 4x².
Replacing x with 2x compresses the graph horizontally by a factor of 2. The equivalent equation 4x² also represents a vertical stretch by a factor of 4.
Answer: D
Hard Mixed Questions · Hard · Question 62
A quadratic function has zeros at x = 2 and x = 6 and a leading coefficient of 1. What is the value of the constant term?
A) 8 B) 10 C) 12 D) 16
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Solution: The function is f(x) = (x − 2)(x − 6).
Expanding gives f(x) = x² − 8x + 12, so the constant term is 12.
Answer: C
Hard Mixed Questions · Hard · Question 63
The line y = 2x + k is tangent to the parabola y = x² − 4x + 7. What is the value of k?
A) −4 B) −2 C) 2 D) 4
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Solution: Set the equations equal: x² − 4x + 7 = 2x + k.
This gives x² − 6x + (7 − k) = 0. Tangency means exactly one solution, so the discriminant is 0.
36 − 4(7 − k) = 0 → 8 + 4k = 0 → k = −2.
Answer: B
Hard Mixed Questions · Hard · Question 64
The height of a ball is modeled by h(t) = −5t² + 30t + 2, where t is measured in seconds. What is the maximum height?
A) 32 B) 42 C) 47 D) 52
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Solution: The maximum occurs at t = −b/(2a) = −30/[2(−5)] = 3.
h(3) = −5(9) + 30(3) + 2 = −45 + 90 + 2 = 47.
Answer: C
Hard Mixed Questions · Hard · Question 65
The equation x² + kx + 16 = 0 has exactly one real solution. What is the value of k²?
A) 16 B) 32 C) 64 D) 128
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Solution: Exactly one real solution means the discriminant is 0.
k² − 4(1)(16) = 0 → k² − 64 = 0 → k² = 64.
Answer: C
What Mistakes Cost Students Points on Parabolas?
Most parabola errors come from reading the form incorrectly, answering with the wrong coordinate, or applying the correct formula without interpreting what the question asks. Use this checklist during review.
| Common Mistake | Why It Causes an Error | How to Fix It |
| Reading (x + 4)² as h = 4 | The sign inside vertex form is reversed. | Read x + 4 as x − (−4), so h = −4. |
| Giving the x-coordinate instead of the maximum or minimum value | The question asks for a y-value, but the student reports when it occurs. | Underline whether the question asks “when” or “what value.” |
| Using −b/(2a) and stopping | The formula gives only the x-coordinate of the vertex. | Substitute the x-value back into the function to find y. |
| Ignoring the discriminant | Students guess the number of intersections from an equation. | Use b² − 4ac to determine 0, 1, or 2 real solutions. |
| Using roots when the vertex is needed | Different features answer different real-world questions. | Use roots for break-even or ground time; use the vertex for a maximum or minimum. |
How Should You Study SAT Parabolas in 2 Weeks?
A focused two-week plan should move from form recognition to timed mixed practice. Review every missed question by writing the feature you should have identified first.
| Time | Focus | What to Do |
| Days 1–2 | Vertex form | Read the vertex, axis of symmetry, opening direction, and maximum or minimum. |
| Days 3–4 | Standard form | Find the y-intercept, vertex, roots, and discriminant. |
| Days 5–6 | Word problems | Practice projectile motion, revenue, area, and optimization models. |
| Day 7 | Transformations | Convert between standard, factored, and vertex forms. |
| Days 8–9 | Hard mixed questions | Practice parameters, systems, tangency, and function notation. |
| Day 10 | Timed set | Complete 15–20 parabola questions without notes. |
| Days 11–12 | Error correction | Redo missed questions and record the exact trap. |
| Day 13 | Mixed SAT Math module | Practice parabolas inside a broader timed module. |
| Day 14 | Final review | Review formulas, error patterns, and five difficult questions. |
Build a Stronger SAT Math Plan
Combine these parabola questions with guided error review, topic-wise practice, and a personalized SAT study plan.
Frequently Asked Questions About SAT Parabolas
Are parabola questions tested on the Digital SAT?
Yes. Parabolas are tested mainly within the Advanced Math domain through quadratic functions, graphs, vertex form, roots, transformations, and real-world models.
What is the fastest way to find the vertex of a parabola?
In vertex form a(x − h)² + k, read the vertex directly as (h, k). In standard form ax² + bx + c, first calculate x = −b/(2a), then substitute that x-value into the function.
Can Desmos solve SAT parabola questions?
Desmos can graph equations and show intersections, roots, and vertices, but you still need to interpret what each point or coefficient means. Algebra is often faster for direct form-reading questions.
What is the difference between a root and a vertex?
A root is an x-value where the parabola crosses or touches the x-axis. The vertex is the maximum or minimum point and lies on the axis of symmetry.
How many parabola questions should I practice before the SAT?
Most students benefit from 50–70 mixed questions covering vertex form, standard form, intercepts, transformations, modeling, discriminants, and hard parameter problems.
He is a Digital SAT mentor with 10+ years of experience, working primarily with SAT students all Over worldwide. Their students have consistently progressed toward 1520+ scores by improving timing, accuracy, and trap-answer control through official-style practice, detailed mistake analysis, and clear weekly action plans.
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