Question

graph the function $f(x) = -5(x + 7)^2 + 10$. plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Explanation:

Step1: Identify vertex from vertex form

The function is in vertex form $f(x)=a(x-h)^2+k$, where vertex is $(h,k)$. For $f(x)=-5(x+7)^2+10$, rewrite $x+7$ as $x-(-7)$, so $h=-7$, $k=10$. Vertex: $(-7, 10)$

Step2: Find a second point

Choose $x=-6$, substitute into $f(x)$:
$f(-6)=-5(-6+7)^2+10 = -5(1)^2+10 = -5+10=5$. Point: $(-6,5)$

Step3: Reflect for symmetric point

Parabola is symmetric about $x=-7$. The point $(-6,5)$ is 1 unit right of vertex, so symmetric point is 1 unit left: $x=-8$, $f(-8)=-5(-8+7)^2+10=-5(1)+10=5$. Point: $(-8,5)$

Answer:

1. Plot the vertex at $(-7, 10)$
2. Plot a second point at $(-6, 5)$ (or $(-8, 5)$)
3. Draw a downward-opening parabola through these points (since $a=-5<0$, the parabola opens downward)