Question

graph the parabola.
y = x² + 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 the vertex of the parabola

For a quadratic function \( y = ax^2+bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y=x^{2}+10x + 21 \), \( a = 1 \), \( b=10 \), \( c = 21 \). So \( x=-\frac{10}{2\times1}=- 5 \).
Substitute \( x=-5 \) into the function: \( y=(-5)^{2}+10\times(-5)+21=25 - 50 + 21=-4 \). So the vertex is \( (-5,-4) \).

Step2: Find points to the left of the vertex

Let's choose \( x=-6 \) (left of \( x = - 5 \)):
\( y=(-6)^{2}+10\times(-6)+21=36-60 + 21=-3 \). So the point is \( (-6,-3) \).
Let's choose \( x=-7 \) (left of \( x=-5 \)):
\( y=(-7)^{2}+10\times(-7)+21=49-70 + 21=0 \). So the point is \( (-7,0) \).

Step3: Find points to the right of the vertex

Let's choose \( x=-4 \) (right of \( x=-5 \)):
\( y=(-4)^{2}+10\times(-4)+21=16-40 + 21=-3 \). So the point is \( (-4,-3) \).
Let's choose \( x=-3 \) (right of \( x=-5 \)):
\( y=(-3)^{2}+10\times(-3)+21=9-30 + 21=0 \). So the point is \( (-3,0) \).

Answer:

The vertex is \((-5, - 4)\), two points to the left of the vertex are \((-6,-3)\) and \((-7,0)\), and two points to the right of the vertex are \((-4,-3)\) and \((-3,0)\). These points can be plotted to graph the parabola \(y=x^{2}+10x + 21\).