Math 2 Quiz: Quadratic Function Features
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Quadratic Function FeaturesQuestion 1 of 11

A parabola has the equation y=2x2−12x+10y = 2x^2 - 12x + 10. If this parabola is shifted 3 units to the right and 2 units down, what is the xx-coordinate of the new vertex?

00
33
99
66
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Math 2 Quiz

Math 2 Quiz: Quadratic Function Features

Practice Quadratic Function Features in Math 2 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 Quadratic Function Features, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

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.

All questions

Question 1

A parabola has the equation y=2x2−12x+10y = 2x^2 - 12x + 10. If this parabola is shifted 3 units to the right and 2 units down, what is the xx-coordinate of the new vertex?

  1. 00
  2. 33
  3. 99
  4. 66 (correct answer)
Explanation: When you encounter parabola transformations, you need to find the original vertex first, then apply the shifts to determine the new vertex location. To find the vertex of y=2x2−12x+10y = 2x^2 - 12x + 10, complete the square or use the vertex formula. The xx-coordinate of the vertex is x=−b2a=−−122(2)=124=3x = -\frac{b}{2a} = -\frac{-12}{2(2)} = \frac{12}{4} = 3. So the original vertex is at (3,−8)(3, -8). When a parabola shifts 3 units right and 2 units down, you add 3 to the xx-coordinate and subtract 2 from the yy-coordinate of the vertex. The new vertex becomes (3+3,−8−2)=(6,−10)(3 + 3, -8 - 2) = (6, -10). Therefore, the xx-coordinate of the new vertex is 6, making answer D correct. Let's examine why the other answers are wrong. Answer A (00) might result from incorrectly subtracting the shift instead of adding it: 3−3=03 - 3 = 0. Answer B (33) is the xx-coordinate of the original vertex before any transformation—this represents forgetting to apply the horizontal shift entirely. Answer C (99) could come from mistakenly multiplying the original xx-coordinate by the shift: 3×3=93 \times 3 = 9, rather than adding them. Remember that horizontal shifts always affect the xx-coordinate by addition (right shift = add, left shift = subtract), while vertical shifts affect the yy-coordinate. Always find the original vertex first, then systematically apply each transformation to avoid mixing up the coordinates or operations.

Question 2

A quadratic function has its vertex at (2,−8)(2, -8) and passes through the point (0,−4)(0, -4). What is the yy-intercept of this function?

  1. −4-4 (correct answer)
  2. −8-8
  3. 44
  4. 88
Explanation: The y-intercept occurs when x = 0. We're told the function passes through (0, -4), so the y-intercept is -4. Choice B incorrectly uses the y-coordinate of the vertex. Choice C uses the opposite sign of the correct answer. Choice D uses the opposite sign of the vertex's y-coordinate.

Question 3

A quadratic function has x-intercepts at x=−3x = -3 and x=7x = 7, and its vertex has a y-coordinate of −25-25. What is the y-intercept of this function?

  1. −21-21 (correct answer)
  2. 2121
  3. −15-15
  4. 2525
Explanation: With x-intercepts at -3 and 7, the function can be written as f(x)=a(x+3)(x−7)f(x) = a(x + 3)(x - 7). The vertex x-coordinate is at x=−3+72=2x = \frac{-3 + 7}{2} = 2. Since the vertex y-coordinate is -25, we have f(2)=a(2+3)(2−7)=a(5)(−5)=−25a=−25f(2) = a(2 + 3)(2 - 7) = a(5)(-5) = -25a = -25, so a=1a = 1. Therefore, f(x)=(x+3)(x−7)=x2−4x−21f(x) = (x + 3)(x - 7) = x^2 - 4x - 21. The y-intercept is f(0)=−21f(0) = -21. Choice B incorrectly uses positive 21. Choice C miscalculates the constant term. Choice D confuses the y-intercept with the vertex y-coordinate's absolute value.

Question 4

The quadratic function g(x)=2x2−8x+kg(x) = 2x^2 - 8x + k has its vertex on the x-axis. If the function is reflected across the x-axis and then shifted up 3 units, what is the y-coordinate of the vertex of the resulting function?

  1. 00
  2. 33 (correct answer)
  3. −3-3
  4. 66
Explanation: First, find k. The vertex of g(x)=2x2−8x+kg(x) = 2x^2 - 8x + k occurs at x=84=2x = \frac{8}{4} = 2. The vertex y-coordinate is g(2)=2(4)−8(2)+k=8−16+k=k−8g(2) = 2(4) - 8(2) + k = 8 - 16 + k = k - 8. Since the vertex is on the x-axis, k−8=0k - 8 = 0, so k=8k = 8. The original vertex is at (2, 0). Reflecting across the x-axis gives vertex (2, 0), then shifting up 3 units gives vertex (2, 3). Choice A forgets the upward shift. Choice C applies the reflection incorrectly after the shift. Choice D doubles the shift amount.

Question 5

A quadratic function f(x)f(x) has its maximum value of 2525 when x=−2x = -2. If f(−5)=7f(-5) = 7, what is the value of f(1)f(1)?

  1. 1616
  2. 2525
  3. 77 (correct answer)
  4. −11-11
Explanation: When you encounter a quadratic function with a given maximum or minimum, you're working with the vertex form of a parabola. Since this function has a maximum value of 25 at x=−2x = -2, you know the vertex is at (−2,25)(-2, 25) and the parabola opens downward. The vertex form is f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex. Substituting the vertex (−2,25)(-2, 25): f(x)=a(x+2)2+25f(x) = a(x + 2)^2 + 25. To find the value of aa, use the given point f(−5)=7f(-5) = 7: 7=a(−5+2)2+257 = a(-5 + 2)^2 + 25 7=a(−3)2+257 = a(-3)^2 + 25 7=9a+257 = 9a + 25 −18=9a-18 = 9a a=−2a = -2 So the complete function is f(x)=−2(x+2)2+25f(x) = -2(x + 2)^2 + 25. Now calculate f(1)f(1): f(1)=−2(1+2)2+25=−2(3)2+25=−2(9)+25=−18+25=7f(1) = -2(1 + 2)^2 + 25 = -2(3)^2 + 25 = -2(9) + 25 = -18 + 25 = 7 The answer is C) 77. Looking at the wrong answers: A) 1616 might result from calculation errors with the coefficient. B) 2525 is the maximum value at the vertex, not the value at x=1x = 1. D) −11-11 could come from sign errors when working with the negative coefficient. Strategy tip: For vertex form problems, always identify the vertex first, then use any additional point to find the coefficient aa. Remember that points equidistant from the vertex have the same yy-value due to the parabola's symmetry.

Question 6

The function h(x)=3(x+4)2−12h(x) = 3(x + 4)^2 - 12 is stretched vertically by a factor of 2, then shifted left 1 unit. What are the coordinates of the vertex of the resulting function?

  1. (−3,−24)(-3, -24)
  2. (−5,−24)(-5, -24) (correct answer)
  3. (−5,−12)(-5, -12)
  4. (−3,−12)(-3, -12)
Explanation: When you encounter function transformations, you need to apply them step-by-step in the order given, tracking how each affects the vertex. Start by identifying the vertex of the original function h(x)=3(x+4)2−12h(x) = 3(x + 4)^2 - 12. This is in vertex form a(x−h)2+ka(x - h)^2 + k, where the vertex is (h,k)(h, k). Since we have (x+4)2(x + 4)^2, that's (x−(−4))2(x - (-4))^2, so the vertex is (−4,−12)(-4, -12). Now apply the transformations in order: Step 1: Vertical stretch by factor of 2 This multiplies the entire function by 2: 2h(x)=2[3(x+4)2−12]=6(x+4)2−242h(x) = 2[3(x + 4)^2 - 12] = 6(x + 4)^2 - 24 The vertex becomes (−4,−24)(-4, -24) because both the aa-value and kk-value are doubled. Step 2: Shift left 1 unit Replace xx with (x+1)(x + 1): 6((x+1)+4)2−24=6(x+5)2−246((x + 1) + 4)^2 - 24 = 6(x + 5)^2 - 24 The vertex moves from (−4,−24)(-4, -24) to (−5,−24)(-5, -24). The answer is (B) (−5,−24)(-5, -24). Why other answers are wrong:
  • (A) (−3,−24)(-3, -24): Incorrectly shifts right instead of left
  • (C) (−5,−12)(-5, -12): Correctly shifts horizontally but forgets the vertical stretch affects the yy-coordinate
  • (D) (−3,−12)(-3, -12): Makes both mistakes above
Study tip: Always apply transformations in the given order, and remember that vertical stretches affect the yy-coordinate of the vertex, while horizontal shifts affect the xx-coordinate.

Question 7

The quadratic function g(x)=−x2+6x−5g(x) = -x^2 + 6x - 5 is reflected across the xx-axis. What is the yy-coordinate of the vertex of the resulting function?

  1. −5-5
  2. 44
  3. −4-4 (correct answer)
  4. 55
Explanation: When you encounter a problem involving reflections of quadratic functions, you need to understand how transformations affect the vertex and overall shape of the parabola. First, let's find the vertex of the original function g(x)=−x2+6x−5g(x) = -x^2 + 6x - 5. For a quadratic in the form ax2+bx+cax^2 + bx + c, the x-coordinate of the vertex is x=−b2a=−62(−1)=3x = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3. Substituting back: g(3)=−(3)2+6(3)−5=−9+18−5=4g(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4. So the original vertex is at (3,4)(3, 4). When a function is reflected across the x-axis, you multiply the entire function by −1-1. This gives us the new function: h(x)=−g(x)=−(−x2+6x−5)=x2−6x+5h(x) = -g(x) = -(-x^2 + 6x - 5) = x^2 - 6x + 5. The reflection flips the parabola vertically, changing the sign of the y-coordinate of every point, including the vertex. Therefore, the new vertex is at (3,−4)(3, -4), making the y-coordinate −4-4, which is choice C. Looking at the wrong answers: A) −5-5 is the original constant term with a sign change, but this isn't how vertex coordinates work. B) 44 is the y-coordinate of the original vertex before reflection—a common trap if you forget to apply the transformation. D) 55 appears to be the original constant term without the sign change. Remember: reflections across the x-axis always change the sign of y-coordinates while keeping x-coordinates unchanged. Find the vertex first, then apply the transformation.

Question 8

A quadratic function has xx-intercepts at x=−6x = -6 and x=2x = 2, and its vertex has yy-coordinate −16-16. What is the yy-intercept of this function?

  1. 1212
  2. −12-12 (correct answer)
  3. −16-16
  4. 1616
Explanation: When you encounter a quadratic function problem with given intercepts and vertex information, you're working with multiple forms of quadratic equations. The most efficient approach is to use the intercept form and then apply the vertex condition. Since the xx-intercepts are at x=−6x = -6 and x=2x = 2, you can write the function as f(x)=a(x+6)(x−2)f(x) = a(x + 6)(x - 2) for some constant aa. The vertex occurs at the midpoint of the intercepts: x=−6+22=−2x = \frac{-6 + 2}{2} = -2. Since the vertex has yy-coordinate −16-16, you know f(−2)=−16f(-2) = -16. Substituting: a(−2+6)(−2−2)=a(4)(−4)=−16a=−16a(-2 + 6)(-2 - 2) = a(4)(-4) = -16a = -16. Therefore, a=1a = 1. The complete function is f(x)=(x+6)(x−2)=x2+4x−12f(x) = (x + 6)(x - 2) = x^2 + 4x - 12. The yy-intercept occurs when x=0x = 0: f(0)=−12f(0) = -12. Looking at the wrong answers: Choice A (1212) gives you the opposite sign of the correct answer—this happens if you incorrectly write the intercept form or make a sign error in expansion. Choice C (−16-16) confuses the yy-coordinate of the vertex with the yy-intercept—these are completely different points on the parabola. Choice D (1616) combines both errors: wrong sign and confusion between vertex and intercept. Study tip: Always distinguish between the vertex (extremum point) and intercepts (axis-crossing points). When using intercept form a(x−r1)(x−r2)a(x - r_1)(x - r_2), remember the vertex xx-coordinate is always r1+r22\frac{r_1 + r_2}{2}.

Question 9

A quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c has the property that f(2)=f(−4)f(2) = f(-4). If the range of the function is (−∞,12](-\infty, 12], what is the axis of symmetry of the function?

  1. x=3x = 3
  2. x=1x = 1
  3. x=−3x = -3
  4. x=−1x = -1 (correct answer)
Explanation: When you see a quadratic function with equal function values at two different x-values, you're looking at symmetry. The key insight is that a parabola is symmetric about its axis of symmetry, so if f(2)=f(−4)f(2) = f(-4), these two points are equidistant from the axis. The axis of symmetry is always at the midpoint between any two points where the function has equal values. To find this midpoint, calculate: x=2+(−4)2=−22=−1x = \frac{2 + (-4)}{2} = \frac{-2}{2} = -1 You can verify this using the range information. Since the range is (−∞,12](-\infty, 12], the maximum value is 12, which occurs at the vertex. For a downward-opening parabola (indicated by the maximum), the vertex is at x=−1x = -1, confirming our axis of symmetry. Looking at the wrong answers: Choice A (x=3x = 3) would make the distance from 2 to the axis equal 1, but the distance from -4 to the axis would be 7 - these aren't equal, so f(2)≠f(−4)f(2) \neq f(-4). Choice B (x=1x = 1) creates distances of 1 and 5 respectively, again unequal. Choice C (x=−3x = -3) gives distances of 5 and 1, which are unequal but in reverse order from choice B. Remember this pattern: when a quadratic has equal function values at two x-coordinates, the axis of symmetry is always the average of those x-coordinates. This saves time and eliminates the need to find the full equation.

Question 10

For the quadratic function m(x)=3x2−12x+7m(x) = 3x^2 - 12x + 7, which statement correctly describes the relationship between the axis of symmetry and the x-intercepts?

  1. The axis of symmetry is x=2x = 2 and the function has no real x-intercepts
  2. The axis of symmetry is x=2x = 2 and the function has two real x-intercepts (correct answer)
  3. The axis of symmetry is x=−2x = -2 and the function has two real x-intercepts
  4. The axis of symmetry is x=−2x = -2 and the function has no real x-intercepts
Explanation: The axis of symmetry is x=−b2a=−−122(3)=126=2x = -\frac{b}{2a} = -\frac{-12}{2(3)} = \frac{12}{6} = 2. To find x-intercepts, solve 3x2−12x+7=03x^2 - 12x + 7 = 0. The discriminant is b2−4ac=(−12)2−4(3)(7)=144−84=60>0b^2 - 4ac = (-12)^2 - 4(3)(7) = 144 - 84 = 60 > 0, so there are two real x-intercepts. Choice A has the correct axis but wrong number of intercepts. Choice C uses the wrong sign for the axis. Choice D has both the wrong axis and wrong number of intercepts.

Question 11

A quadratic function has x-intercepts at (−1,0)(-1, 0) and (5,0)(5, 0), and passes through the point (3,−8)(3, -8). What is the y-coordinate of the vertex?

  1. 99
  2. −12-12
  3. −9-9 (correct answer)
  4. 1212
Explanation: When you encounter a quadratic function problem with given x-intercepts and an additional point, you're working with the factored form of a quadratic. This approach is often the most efficient path to finding vertex information. Since the x-intercepts are at (−1,0)(-1, 0) and (5,0)(5, 0), you can write the quadratic as f(x)=a(x+1)(x−5)f(x) = a(x + 1)(x - 5), where aa is a constant you need to determine. Use the given point (3,−8)(3, -8) to find aa: −8=a(3+1)(3−5)=a(4)(−2)=−8a-8 = a(3 + 1)(3 - 5) = a(4)(-2) = -8a. Therefore, a=1a = 1. Your function is f(x)=(x+1)(x−5)=x2−4x−5f(x) = (x + 1)(x - 5) = x^2 - 4x - 5. The vertex's x-coordinate lies halfway between the x-intercepts: x=−1+52=2x = \frac{-1 + 5}{2} = 2. Substituting: f(2)=22−4(2)−5=4−8−5=−9f(2) = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9. Looking at the wrong answers: Choice A (99) likely comes from forgetting the negative sign when calculating f(2)f(2). Choice B (−12-12) might result from incorrectly using the vertex formula or making an arithmetic error with the given point. Choice D (1212) could stem from sign errors in multiple steps or confusing the direction of the parabola. The correct answer is C: −9-9. Remember this pattern: when you have x-intercepts, use factored form first to find the coefficient, then locate the vertex at the midpoint of the intercepts. This method is typically faster and less error-prone than expanding to standard form immediately.