We have to find 5 points of the parabola:
[tex]y=x^2+8x+11[/tex]
and then graph it.
We can find the vertex by completing the square:
[tex]\begin{gathered} y=x^2+8x+11 \\ y=x^2+2\cdot4x+16-16+11 \\ y=(x+4)^2-5 \end{gathered}[/tex]
As we now have the vertex form of the parabola, we can see that the vertex is at (x,y) = (-4,-5).
We can now calculate two points to the right of the parabola by giving values to x as x = 0 and x = -2:
[tex]y=0^2+8\cdot0+11=11[/tex][tex]\begin{gathered} y=(-2)^2+8\cdot(-2)+11 \\ y=4-16+11 \\ y=-1 \end{gathered}[/tex]
We now know two points to the right of the parabola: (0, 11) and (-2, -1).
As the line x = -4 is the axis of symmetry, we will have the same value for y when the values of x are at the same distance from this line.
Then, we can write:
[tex]\begin{gathered} y(0)=y(-8)=11 \\ y(-2)=y(-6)=-1 \end{gathered}[/tex]
Then, we have two points to the left: (-8, 11) and (-6, -1).
We can graph the parabola as:
