Answer:
See the attached image.
Step-by-step explanation:
To graph the quadratic equation, we can solve for 5 points on the parabola, choosing x-values centered at its vertex.
First, we need to put the quadratic in vertex form by completing the square:
[tex]y=(x^2+10x)+16[/tex]
↓ adding (10/2)² inside the parentheses and subtracting it outside
[tex]y = (x^2+10x\ \underline{+\ 25})+16 \ \underline{-\ 25}[/tex]
↓ factoring the expression in parentheses as a perfect square
[tex]y = (x+5)^2 - 9[/tex]
Now, we have the equation in vertex form:
[tex]y=(x-a)^2 + b[/tex]
where:
- [tex](a,b)[/tex] is the vertex of the parabola.
This means that the parabola's vertex is:
So, to get useful (graphable) y-values, we should use x-values close to -5:
[tex]\begin{array}{| c | c |} \cline{1-2} x & y \\ \cline{1-2} -7 & -5 \\ \cline{1-2} -6 & -8 \\ \cline{1-2} -5 & -9 \\ \cline{1-2} -4 & -8 \\ \cline{1-2} -3 & -5 \\ \cline{1-2} \end{array}[/tex]
From these x- and y-coordinates, we can see that the function is symmetric around the vertex, which makes sense because it is a parabola.
If we graph these five points, it should look something like the attached image.
