Graphing Parabolas: Vertex, Symmetry, Intercepts - College Algebra

1. Consider the parabola $f(x) = -x^2 + 4x + 1$. If point $P$ is the vertex and point $Q$ is the $y$-intercept, what is the slope of line segment $\overline{PQ}$?
   Answer: $2$ 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 $\frac{5 - 1}{2 - 0} = \frac{4}{2} = 2$.
2. A quadratic function has the form $f(x) = 2x^2 + bx + c$ where the axis of symmetry is $x = 3$. If the function has a $y$-intercept of $-10$, what is the $x$-coordinate of the other point on the parabola that has the same $y$-value as the $y$-intercept?
   Answer: $6$ 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: $f(6) = 2(36) - 12(6) - 10 = 72 - 72 - 10 = -10$.
3. The graph of $y = ax^2 + bx + c$ has vertex at $(2, -3)$ and $y$-intercept at $(0, 5)$. What is the value of the coefficient $a$?
   Answer: $2$ 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 $a = 2$.
4. The quadratic function $f(x) = -2x^2 + 8x - 5$ has vertex at point $(h, k)$. If the parabola is reflected across its axis of symmetry, what is the $y$-coordinate of the point on the reflected parabola that corresponds to the original point $(1, 1)$?
   Answer: $1$ 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.
5. The parabola $y = ax^2 + bx + c$ has vertex at $(-2, 7)$ and passes through the point $(1, -2)$. What is the sum of the $x$-coordinates of the $x$-intercepts?
   Answer: $-4$ 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.

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