College Algebra Quiz: Graphing Parabolas Vertex Symmetry Intercepts

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.

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.

Question 1

1. $3$
2. $4$
3. $6$ (correct answer)
4. $8$

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

Question 2

1. $1$ (correct answer)
2. $3$
3. $5$
4. $7$

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

Question 3

1. $-2$
2. $-4$ (correct answer)
3. $0$
4. $4$

Explanation: When you encounter a parabola problem with vertex and point information, remember that the vertex form $y = a(x - h)^2 + k$ is your most efficient tool, where $(h, k)$ is the vertex. Since the vertex is $(-2, 7)$, we have $y = a(x + 2)^2 + 7$. To find $a$, substitute the given point $(1, -2)$: $-2 = a(1 + 2)^2 + 7$ $-2 = 9a + 7$ $-9 = 9a$ $a = -1$ So our parabola is $y = -(x + 2)^2 + 7$. Here's the key insight: for any parabola, the sum of the x-intercepts equals $-\frac{b}{a}$ when written in standard form $y = ax^2 + bx + c$. Let's expand our equation: $y = -(x^2 + 4x + 4) + 7 = -x^2 - 4x + 3$ Therefore, $a = -1$ and $b = -4$, giving us $-\frac{b}{a} = -\frac{(-4)}{(-1)} = -4$. Choice A ($-2$) is the x-coordinate of the vertex, which students often confuse with the sum of roots. Choice C ($0$) would occur if the parabola were symmetric about the y-axis, which isn't the case here. Choice D ($4$) is the positive version of our answer, a common sign error when applying the sum formula. The correct answer is B ($-4$). Study tip: Remember that for parabolas, the sum of x-intercepts always equals $-\frac{b}{a}$, and the x-coordinate of the vertex equals half this sum. This relationship can save you time on similar problems.

Question 4

1. $-\frac{1}{2}$
2. $\frac{1}{2}$
3. $-2$
4. $2$ (correct answer)

Explanation: First, find the vertex. For $f(x) = -x^2 + 4x + 1$, the $x$-coordinate of the vertex is $h = -\frac{4}{2(-1)} = 2$. The $y$-coordinate is $k = f(2) = -4 + 8 + 1 = 5$. So the vertex $P$ is at $(2, 5)$. The $y$-intercept $Q$ is at $(0, f(0)) = (0, 1)$. The slope of line segment $\overline{PQ}$ is $\frac{5 - 1}{2 - 0} = \frac{4}{2} = 2$.

Question 5

1. $\frac{1}{2}$
2. $1$
3. $2$ (correct answer)
4. $4$

Explanation: Using the vertex form $y = a(x - h)^2 + k$ with vertex $(2, -3)$: $y = a(x - 2)^2 - 3$. The $y$-intercept gives us the point $(0, 5)$, so substituting: $5 = a(0 - 2)^2 - 3 = 4a - 3$. Solving for $a$: $5 + 3 = 4a$, so $8 = 4a$, which gives $a = 2$.