Question

I need quick answers please, is due soon. i need assistance finding 5 points. 2 to the left of vertex, i need the vertex, and 2 to the right of the vertex. the graph only goes up to 14. thank you!

Answer 1

We have to find 5 points of the parabola:

`y=x^2+8x+11`

and then graph it.

We can find the vertex by completing the square:

`\begin{gathered} y=x^2+8x+11 \\ y=x^2+2\cdot4x+16-16+11 \\ y=(x+4)^2-5 \end{gathered}`

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:

`y=0^2+8\cdot0+11=11`
`\begin{gathered} y=(-2)^2+8\cdot(-2)+11 \\ y=4-16+11 \\ y=-1 \end{gathered}`

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:

`\begin{gathered} y(0)=y(-8)=11 \\ y(-2)=y(-6)=-1 \end{gathered}`

Then, we have two points to the left: (-8, 11) and (-6, -1).

We can graph the parabola as: