ACCUPLACER Advanced Algebra & Functions Quiz: Graphing Quadratic Functions
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Graphing Quadratic FunctionsQuestion 1 of 20

What is the sum of the coordinates of the vertex of the parabola defined by the function f(x)=−x2−8x−13f(x) = -x^2 - 8x - 13?

-11
-4
-1
3
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Graphing Quadratic Functions

Practice Graphing Quadratic Functions in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Graphing Quadratic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

What is the sum of the coordinates of the vertex of the parabola defined by the function f(x)=−x2−8x−13f(x) = -x^2 - 8x - 13?

  1. -11
  2. -4
  3. -1 (correct answer)
  4. 3
Explanation: First, find the coordinates of the vertex (h,k)(h, k). The x-coordinate is h=−b2a=−−82(−1)=8−2=−4h = -\frac{b}{2a} = -\frac{-8}{2(-1)} = \frac{8}{-2} = -4. The y-coordinate is k=f(−4)=−(−4)2−8(−4)−13=−16+32−13=3k = f(-4) = -(-4)^2 - 8(-4) - 13 = -16 + 32 - 13 = 3. The vertex is (−4,3)(-4, 3). The sum of the coordinates is −4+3=−1-4 + 3 = -1.

Question 2

The parabola given by the equation y=−2x2+bx−9y = -2x^2 + bx - 9 has its axis of symmetry at x=−3x = -3. What is the value of bb?

  1. -12 (correct answer)
  2. -6
  3. 6
  4. 12
Explanation: The formula for the axis of symmetry of a parabola y=ax2+bx+cy = ax^2 + bx + c is x=−b2ax = -\frac{b}{2a}. We are given that the axis of symmetry is x=−3x = -3, and from the equation we have a=−2a = -2. Plugging these values into the formula gives −3=−b2(−2)-3 = -\frac{b}{2(-2)}. This simplifies to −3=−b−4-3 = -\frac{b}{-4}, which is −3=b4-3 = \frac{b}{4}. Multiplying both sides by 4 gives b=−12b = -12.

Question 3

The graph of a quadratic function has a vertex at (3,−8)(3, -8) and a y-intercept at (0,10)(0, 10). What is the leading coefficient of the function?

  1. -2
  2. 1/2
  3. 2 (correct answer)
  4. 10
Explanation: Using the vertex form of a quadratic function, y=a(x−h)2+ky = a(x-h)^2 + k, we can substitute the vertex coordinates (h,k)=(3,−8)(h,k) = (3,-8) to get y=a(x−3)2−8y = a(x-3)^2 - 8. To find the leading coefficient aa, we use the y-intercept (0,10)(0, 10). Substitute x=0x=0 and y=10y=10 into the equation: 10=a(0−3)2−810 = a(0-3)^2 - 8. This simplifies to 10=a(−3)2−810 = a(-3)^2 - 8, or 10=9a−810 = 9a - 8. Adding 8 to both sides gives 18=9a18 = 9a. Dividing by 9 gives a=2a = 2.

Question 4

What is the distance between the x-intercepts of the function f(x)=3(x−4)2−48f(x) = 3(x-4)^2 - 48?

  1. 0
  2. 4
  3. 8 (correct answer)
  4. 16
Explanation: To find the x-intercepts, set f(x)=0f(x) = 0: 3(x−4)2−48=03(x-4)^2 - 48 = 0. Add 48 to both sides: 3(x−4)2=483(x-4)^2 = 48. Divide by 3: (x−4)2=16(x-4)^2 = 16. Take the square root of both sides: x−4=±4x-4 = \pm 4. This gives two equations: x−4=4x-4 = 4, which solves to x=8x = 8, and x−4=−4x-4 = -4, which solves to x=0x = 0. The two x-intercepts are 8 and 0. The distance between them is ∣8−0∣=8|8 - 0| = 8.

Question 5

The graph of the quadratic function g(x)=−2x2−20x−40g(x) = -2x^2 - 20x - 40 has a maximum value. What is this maximum value?

  1. -40
  2. -5
  3. 10 (correct answer)
  4. 40
Explanation: The maximum value of a downward-opening parabola is the y-coordinate of its vertex. First, find the x-coordinate of the vertex: h=−b2a=−−202(−2)=20−4=−5h = -\frac{b}{2a} = -\frac{-20}{2(-2)} = \frac{20}{-4} = -5. Then, find the y-coordinate by substituting hh into the function: k=g(−5)=−2(−5)2−20(−5)−40=−2(25)+100−40=−50+100−40=10k = g(-5) = -2(-5)^2 - 20(-5) - 40 = -2(25) + 100 - 40 = -50 + 100 - 40 = 10. The maximum value is 10.

Question 6

The axis of symmetry for a parabola is the line x=5x = 5. One point on the parabola is (2,7)(2, 7). Which of the following points must also be on the parabola?

  1. (-2, 7)
  2. (2, -7)
  3. (5, 7)
  4. (8, 7) (correct answer)
Explanation: Parabolas are symmetric about their axis of symmetry. The point (2,7)(2, 7) has an x-coordinate of 2, which is 5−3=25 - 3 = 2, or 3 units to the left of the axis of symmetry x=5x = 5. Due to symmetry, there must be another point with the same y-coordinate that is 3 units to the right of the axis of symmetry. Its x-coordinate will be 5+3=85 + 3 = 8. Therefore, the point (8,7)(8, 7) must also be on the parabola.

Question 7

The parabola y=x2+bx+cy = x^2 + bx + c has x-intercepts at (2,0)(2,0) and (8,0)(8,0). What are the coordinates of its vertex?

  1. (5, 0)
  2. (5, -9) (correct answer)
  3. (5, 9)
  4. (10, 16)
Explanation: The x-coordinate of the vertex is the average of the x-intercepts: h=2+82=5h = \frac{2+8}{2} = 5. To find the equation, we can use the sum and product of roots. The sum of the roots is 2+8=102+8=10, and for y=x2+bx+cy=x^2+bx+c, the sum is −b-b. So 10=−b10=-b, which means b=−10b=-10. The product of the roots is 2×8=162 \times 8=16, which is equal to cc. The equation is y=x2−10x+16y = x^2 - 10x + 16. Now, find the y-coordinate of the vertex by evaluating the function at x=5x=5: k=(5)2−10(5)+16=25−50+16=−9k = (5)^2 - 10(5) + 16 = 25 - 50 + 16 = -9. The vertex is (5,−9)(5, -9).

Question 8

For the quadratic function f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k, it is known that a>0a>0 and k<0k<0. Which statement accurately describes the x-intercepts of the graph of f(x)f(x)?

  1. There are no x-intercepts.
  2. There is exactly one x-intercept.
  3. There are two distinct x-intercepts. (correct answer)
  4. The number of x-intercepts cannot be determined.
Explanation: The parameter aa determines the direction the parabola opens. Since a>0a > 0, the parabola opens upwards. The vertex of the parabola is at the point (h,k)(h, k). Since k<0k < 0, the vertex is located below the x-axis. A parabola that opens upwards and has its minimum point (the vertex) below the x-axis must cross the x-axis twice. Therefore, there are two distinct x-intercepts.

Question 9

The range of the quadratic function f(x)=−3x2+6x+5f(x) = -3x^2 + 6x + 5 is (−∞,k](-\infty, k]. What is the value of kk?

  1. -4
  2. 1
  3. 5
  4. 8 (correct answer)
Explanation: The function represents a downward-opening parabola because the leading coefficient (-3) is negative. The range is determined by the maximum value of the function, which is the y-coordinate of the vertex. First, find the x-coordinate of the vertex: h=−b2a=−62(−3)=−6−6=1h = -\frac{b}{2a} = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1. Then find the y-coordinate: k=f(1)=−3(1)2+6(1)+5=−3+6+5=8k = f(1) = -3(1)^2 + 6(1) + 5 = -3 + 6 + 5 = 8. Thus, the maximum value is 8, and the range is (−∞,8](-\infty, 8]. The value of kk is 8.

Question 10

The graph of the quadratic function y=2(x+3)2−8y = 2(x+3)^2 - 8 is translated 4 units to the right and 3 units up. What is the vertex of the new, translated parabola?

  1. (-7, -11)
  2. (1, -5) (correct answer)
  3. (-1, -5)
  4. (7, -11)
Explanation: The original function is in vertex form, y=a(x−h)2+ky = a(x-h)^2 + k, where the vertex is (h,k)(h, k). For y=2(x+3)2−8y = 2(x+3)^2 - 8, which is y=2(x−(−3))2−8y = 2(x - (-3))^2 - 8, the vertex is (−3,−8)(-3, -8). Translating 4 units to the right adds 4 to the x-coordinate: −3+4=1-3 + 4 = 1. Translating 3 units up adds 3 to the y-coordinate: −8+3=−5-8 + 3 = -5. The new vertex is (1,−5)(1, -5).

Question 11

A parabola has a y-intercept at (0,−6)(0, -6) and its vertex is at (−2,2)(-2, 2). Which of the following values is one of its x-intercepts?

  1. -3 (correct answer)
  2. -2
  3. 1
  4. 3
Explanation: Start with the vertex form y=a(x−h)2+ky = a(x-h)^2 + k. Substitute the vertex (−2,2)(-2, 2): y=a(x+2)2+2y = a(x+2)^2 + 2. Use the y-intercept (0,−6)(0, -6) to find aa: −6=a(0+2)2+2-6 = a(0+2)^2 + 2, which simplifies to −6=4a+2-6 = 4a + 2, so −8=4a-8 = 4a, and a=−2a = -2. The equation is y=−2(x+2)2+2y = -2(x+2)^2 + 2. To find the x-intercepts, set y=0y=0: 0=−2(x+2)2+20 = -2(x+2)^2 + 2. This gives −2=−2(x+2)2-2 = -2(x+2)^2, so 1=(x+2)21 = (x+2)^2. Taking the square root gives ±1=x+2\pm 1 = x+2. The two solutions are x=−1−2=−3x = -1 - 2 = -3 and x=1−2=−1x = 1 - 2 = -1. One of the x-intercepts is -3.

Question 12

The minimum value of a quadratic function f(x)f(x) is -5. Its axis of symmetry is x=2x = 2. The graph also passes through the point (4,3)(4, 3). What is the y-intercept of the graph?

  1. -5
  2. 2
  3. 3 (correct answer)
  4. 5
Explanation: The minimum value and axis of symmetry give the vertex of the parabola, which is (2,−5)(2, -5). Using the vertex form y=a(x−h)2+ky = a(x-h)^2 + k, we have y=a(x−2)2−5y = a(x-2)^2 - 5. Use the point (4,3)(4, 3) to find the value of aa: 3=a(4−2)2−53 = a(4-2)^2 - 5. This simplifies to 3=a(2)2−53 = a(2)^2 - 5, or 3=4a−53 = 4a - 5. Adding 5 to both sides gives 8=4a8 = 4a, so a=2a=2. The equation is y=2(x−2)2−5y = 2(x-2)^2 - 5. To find the y-intercept, set x=0x=0: y=2(0−2)2−5=2(−2)2−5=2(4)−5=8−5=3y = 2(0-2)^2 - 5 = 2(-2)^2 - 5 = 2(4) - 5 = 8 - 5 = 3. The y-intercept is 3.

Question 13

The graph of a parabola is given by the function f(x)=x2−5x−14f(x) = x^2 - 5x - 14. What is the sum of its y-intercept and its positive x-intercept?

  1. -16
  2. -7 (correct answer)
  3. 5
  4. 9
Explanation: To find the y-intercept, set x=0x=0: f(0)=02−5(0)−14=−14f(0) = 0^2 - 5(0) - 14 = -14. To find the x-intercepts, set f(x)=0f(x)=0: x2−5x−14=0x^2 - 5x - 14 = 0. Factoring the quadratic gives (x−7)(x+2)=0(x-7)(x+2) = 0. The x-intercepts are x=7x=7 and x=−2x=-2. The positive x-intercept is 7. The required sum is the y-intercept plus the positive x-intercept: −14+7=−7-14 + 7 = -7.

Question 14

What is the product of the coordinates of the vertex of the function f(x)=2x2+8x−1f(x) = 2x^2 + 8x - 1?

  1. -18
  2. -11
  3. 11
  4. 18 (correct answer)
Explanation: First, find the coordinates of the vertex (h,k)(h, k). The x-coordinate is h=−b2a=−82(2)=−84=−2h = -\frac{b}{2a} = -\frac{8}{2(2)} = -\frac{8}{4} = -2. The y-coordinate is k=f(−2)=2(−2)2+8(−2)−1=2(4)−16−1=8−16−1=−9k = f(-2) = 2(-2)^2 + 8(-2) - 1 = 2(4) - 16 - 1 = 8 - 16 - 1 = -9. The vertex is (−2,−9)(-2, -9). The product of the coordinates is (−2)×(−9)=18(-2) \times (-9) = 18.

Question 15

The vertex of the parabola y=ax2+16x+2y = ax^2 + 16x + 2 is at x=−4x=-4. What is the y-coordinate of the vertex?

  1. -30 (correct answer)
  2. -4
  3. 2
  4. 34
Explanation: The x-coordinate of the vertex is given by h=−b2ah = -\frac{b}{2a}. We are given h=−4h = -4, b=16b = 16. So, −4=−162a-4 = -\frac{16}{2a}, which simplifies to −4=−8a-4 = -\frac{8}{a}. Solving for aa gives −4a=−8-4a = -8, so a=2a = 2. The function is y=2x2+16x+2y = 2x^2 + 16x + 2. To find the y-coordinate of the vertex, substitute x=−4x = -4 into the function: k=2(−4)2+16(−4)+2=2(16)−64+2=32−64+2=−30k = 2(-4)^2 + 16(-4) + 2 = 2(16) - 64 + 2 = 32 - 64 + 2 = -30.

Question 16

A parabola has x-intercepts at x=−1x = -1 and x=5x = 5, and it passes through the point (1,−16)(1, -16). What is the y-coordinate of its vertex?

  1. -18 (correct answer)
  2. -16
  3. -9
  4. 2
Explanation: The x-coordinate of the vertex is the midpoint of the x-intercepts: h=−1+52=42=2h = \frac{-1 + 5}{2} = \frac{4}{2} = 2. The equation can be written in factored form as y=a(x+1)(x−5)y = a(x+1)(x-5). Use the point (1,−16)(1, -16) to find aa: −16=a(1+1)(1−5)=a(2)(−4)=−8a-16 = a(1+1)(1-5) = a(2)(-4) = -8a, so a=2a=2. The equation is y=2(x+1)(x−5)y = 2(x+1)(x-5). To find the y-coordinate of the vertex, evaluate the function at x=2x=2: k=2(2+1)(2−5)=2(3)(−3)=−18k = 2(2+1)(2-5) = 2(3)(-3) = -18.

Question 17

A parabola has a vertex at (1,9)(1, 9) and one x-intercept at (4,0)(4, 0). What is the x-coordinate of the other x-intercept?

  1. -4
  2. -3
  3. -2 (correct answer)
  4. 2.5
Explanation: The axis of symmetry is a vertical line passing through the vertex, so its equation is x=1x=1. The x-intercepts are symmetric with respect to the axis of symmetry. The given x-intercept, 4, is 4−1=34 - 1 = 3 units to the right of the axis of symmetry. The other x-intercept must be 3 units to the left of the axis of symmetry. Its x-coordinate is 1−3=−21 - 3 = -2.

Question 18

A quadratic function has exactly one x-intercept, which is at x=4x=4. The graph of the function passes through the point (6,8)(6, 8). What are the coordinates of the vertex?

  1. (4, 8)
  2. (4, 0) (correct answer)
  3. (5, 2)
  4. (6, 8)
Explanation: If a quadratic function has exactly one x-intercept, that intercept must be the vertex of the parabola. Therefore, the vertex is located at the point where the graph touches the x-axis, which is (4,0)(4, 0). The information that the graph passes through (6,8)(6, 8) is extra information used to confirm the equation but is not needed to locate the vertex.

Question 19

The graph of the quadratic function f(x)=2x2−12x+11f(x) = 2x^2 - 12x + 11 is a parabola. What is the y-coordinate of the vertex of this parabola?

  1. -7 (correct answer)
  2. 3
  3. 11
  4. 29
Explanation: The x-coordinate of the vertex of a parabola given by f(x)=ax2+bx+cf(x) = ax^2 + bx + c is x=−b2ax = -\frac{b}{2a}. For this function, a=2a=2 and b=−12b=-12, so the x-coordinate is x=−−122(2)=124=3x = -\frac{-12}{2(2)} = \frac{12}{4} = 3. To find the y-coordinate, substitute this x-value back into the function: f(3)=2(3)2−12(3)+11=2(9)−36+11=18−36+11=−7f(3) = 2(3)^2 - 12(3) + 11 = 2(9) - 36 + 11 = 18 - 36 + 11 = -7. Therefore, the y-coordinate of the vertex is -7.

Question 20

The graph of y=−x2+10x+cy = -x^2 + 10x + c has its vertex on the x-axis. What is the value of cc?

  1. -25 (correct answer)
  2. 0
  3. 5
  4. 25
Explanation: If the vertex is on the x-axis, the y-coordinate of the vertex is 0. First, find the x-coordinate of the vertex: h=−b2a=−102(−1)=5h = -\frac{b}{2a} = -\frac{10}{2(-1)} = 5. Now, substitute x=5x=5 into the equation and set y=0y=0 to find cc: 0=−(5)2+10(5)+c0 = -(5)^2 + 10(5) + c. This simplifies to 0=−25+50+c0 = -25 + 50 + c, or 0=25+c0 = 25 + c. Solving for cc gives c=−25c = -25.