College Algebra Quiz: Graphing Parabolas Vertex Symmetry Intercepts
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Graphing Parabolas Vertex Symmetry InterceptsQuestion 1 of 5

A quadratic function has the form f(x)=2x2+bx+cf(x) = 2x^2 + bx + c where the axis of symmetry is x=3x = 3. If the function has a yy-intercept of −10-10, what is the xx-coordinate of the other point on the parabola that has the same yy-value as the yy-intercept?

33
44
66
88
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College Algebra Quiz

College Algebra Quiz: Graphing Parabolas Vertex Symmetry Intercepts

Practice Graphing Parabolas Vertex Symmetry Intercepts in College Algebra 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 Parabolas Vertex Symmetry Intercepts, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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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.

All questions

Question 1

A quadratic function has the form f(x)=2x2+bx+cf(x) = 2x^2 + bx + c where the axis of symmetry is x=3x = 3. If the function has a yy-intercept of −10-10, what is the xx-coordinate of the other point on the parabola that has the same yy-value as the yy-intercept?

  1. 33
  2. 44
  3. 66 (correct answer)
  4. 88
Explanation: Since the axis of symmetry is x=3x = 3, we have −b2(2)=3-\frac{b}{2(2)} = 3, so −b4=3-\frac{b}{4} = 3, giving b=−12b = -12. The yy-intercept gives us f(0)=c=−10f(0) = c = -10. So f(x)=2x2−12x−10f(x) = 2x^2 - 12x - 10. The yy-intercept occurs at (0,−10)(0, -10). Due to symmetry about x=3x = 3, the point (0,−10)(0, -10) is 3 units to the left of the axis of symmetry. The corresponding point with the same yy-value must be 3 units to the right of the axis of symmetry, at x=3+3=6x = 3 + 3 = 6. We can verify: f(6)=2(36)−12(6)−10=72−72−10=−10f(6) = 2(36) - 12(6) - 10 = 72 - 72 - 10 = -10.

Question 2

The quadratic function f(x)=−2x2+8x−5f(x) = -2x^2 + 8x - 5 has vertex at point (h,k)(h, k). If the parabola is reflected across its axis of symmetry, what is the yy-coordinate of the point on the reflected parabola that corresponds to the original point (1,1)(1, 1)?

  1. 11 (correct answer)
  2. 33
  3. 55
  4. 77
Explanation: First, find the vertex by completing the square or using h=−b2a=−82(−2)=2h = -\frac{b}{2a} = -\frac{8}{2(-2)} = 2. The axis of symmetry is x=2x = 2. The point (1,1)(1, 1) is 1 unit to the left of the axis of symmetry. Its reflection across x=2x = 2 is the point (3,y)(3, y) where y=f(3)=−2(3)2+8(3)−5=−18+24−5=1y = f(3) = -2(3)^2 + 8(3) - 5 = -18 + 24 - 5 = 1. Since parabolas are symmetric about their axis, the yy-coordinate remains the same.

Question 3

The parabola y=ax2+bx+cy = ax^2 + bx + c has vertex at (−2,7)(-2, 7) and passes through the point (1,−2)(1, -2). What is the sum of the xx-coordinates of the xx-intercepts?

  1. −2-2
  2. −4-4 (correct answer)
  3. 00
  4. 44
Explanation: When you encounter a parabola problem with vertex and point information, remember that the vertex form y=a(x−h)2+ky = a(x - h)^2 + k is your most efficient tool, where (h,k)(h, k) is the vertex. Since the vertex is (−2,7)(-2, 7), we have y=a(x+2)2+7y = a(x + 2)^2 + 7. To find aa, substitute the given point (1,−2)(1, -2): −2=a(1+2)2+7-2 = a(1 + 2)^2 + 7 −2=9a+7-2 = 9a + 7 −9=9a-9 = 9a a=−1a = -1 So our parabola is y=−(x+2)2+7y = -(x + 2)^2 + 7. Here's the key insight: for any parabola, the sum of the x-intercepts equals −ba-\frac{b}{a} when written in standard form y=ax2+bx+cy = ax^2 + bx + c. Let's expand our equation: y=−(x2+4x+4)+7=−x2−4x+3y = -(x^2 + 4x + 4) + 7 = -x^2 - 4x + 3 Therefore, a=−1a = -1 and b=−4b = -4, giving us −ba=−(−4)(−1)=−4-\frac{b}{a} = -\frac{(-4)}{(-1)} = -4. Choice A (−2-2) is the x-coordinate of the vertex, which students often confuse with the sum of roots. Choice C (00) would occur if the parabola were symmetric about the y-axis, which isn't the case here. Choice D (44) is the positive version of our answer, a common sign error when applying the sum formula. The correct answer is B (−4-4). Study tip: Remember that for parabolas, the sum of x-intercepts always equals −ba-\frac{b}{a}, and the x-coordinate of the vertex equals half this sum. This relationship can save you time on similar problems.

Question 4

Consider the parabola f(x)=−x2+4x+1f(x) = -x^2 + 4x + 1. If point PP is the vertex and point QQ is the yy-intercept, what is the slope of line segment PQ‾\overline{PQ}?

  1. −12-\frac{1}{2}
  2. 12\frac{1}{2}
  3. −2-2
  4. 22 (correct answer)
Explanation: First, find the vertex. For f(x)=−x2+4x+1f(x) = -x^2 + 4x + 1, the xx-coordinate of the vertex is h=−42(−1)=2h = -\frac{4}{2(-1)} = 2. The yy-coordinate is k=f(2)=−4+8+1=5k = f(2) = -4 + 8 + 1 = 5. So the vertex PP is at (2,5)(2, 5). The yy-intercept QQ is at (0,f(0))=(0,1)(0, f(0)) = (0, 1). The slope of line segment PQ‾\overline{PQ} is 5−12−0=42=2\frac{5 - 1}{2 - 0} = \frac{4}{2} = 2.

Question 5

The graph of y=ax2+bx+cy = ax^2 + bx + c has vertex at (2,−3)(2, -3) and yy-intercept at (0,5)(0, 5). What is the value of the coefficient aa?

  1. 12\frac{1}{2}
  2. 11
  3. 22 (correct answer)
  4. 44
Explanation: Using the vertex form y=a(x−h)2+ky = a(x - h)^2 + k with vertex (2,−3)(2, -3): y=a(x−2)2−3y = a(x - 2)^2 - 3. The yy-intercept gives us the point (0,5)(0, 5), so substituting: 5=a(0−2)2−3=4a−35 = a(0 - 2)^2 - 3 = 4a - 3. Solving for aa: 5+3=4a5 + 3 = 4a, so 8=4a8 = 4a, which gives a=2a = 2.