Question

graph the parabola.
$y = x^2 + 10x + 21$
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph - a - function button.

Explanation:

Step1: Find vertex x-coordinate

For $y=ax^2+bx+c$, $x=-\frac{b}{2a}$.
Here $a=1$, $b=10$, so $x=-\frac{10}{2(1)}=-5$

Step2: Find vertex y-coordinate

Substitute $x=-5$ into the equation.
$y=(-5)^2+10(-5)+21=25-50+21=-4$
Vertex: $(-5, -4)$

Step3: Find left points (x=-7, -6)

For $x=-7$: $y=(-7)^2+10(-7)+21=49-70+21=0$
For $x=-6$: $y=(-6)^2+10(-6)+21=36-60+21=-3$
Left points: $(-7, 0)$, $(-6, -3)$

Step4: Find right points (x=-4, -3)

For $x=-4$: $y=(-4)^2+10(-4)+21=16-40+21=-3$
For $x=-3$: $y=(-3)^2+10(-3)+21=9-30+21=0$
Right points: $(-4, -3)$, $(-3, 0)$

Answer:

Plot the following 5 points:

1. Vertex: $(-5, -4)$
2. Left points: $(-7, 0)$, $(-6, -3)$
3. Right points: $(-4, -3)$, $(-3, 0)$

Then connect the points to graph the parabola.