Question

Can you help with the portion "plot five points: 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.Graph the parabola.

Answer 1

Let's begin by listing out the given information:

`\begin{gathered} y=(x+5)^2+1 \\ y=(x+5)(x+5)+1 \\ y=x(x+5)+5(x+5)+1 \\ y=x^2+5x+5x+25+1 \\ y=x^2+10x+26 \\ \\ a=1,b=10,c=26 \end{gathered}`

The vertex of the function is given by:

`\begin{gathered} x=-(b)/(2a) \\ x=-(10)/(2(1))=-(10)/(2) \\ x=-5 \\ \\ y=(x+5)^2+1 \\ y=(-5+5)^2+1=0^2+1 \\ y=1 \\ \\ (x,y)=(h,k)=(-5,1) \\ \end{gathered}`
`\begin{gathered} y=x^2+10x+26 \\ x=-3 \\ y=-3^2+10(-3)+26=5 \\ x=-4 \\ y=-4^2+10(-4)+26=2 \\ x=-6 \\ y=-6^2+10(-6)+26=2 \\ x=-7 \\ y=-7^2+10(-7)+26=5 \\ x=-5 \\ y=-5^2+10(-5)+26=1 \\ \mleft(x,y\mright)=\mleft(-3,5\mright),\mleft(-4,2\mright),(-5,1),\mleft(-6,2\mright),\mleft(-7,5\mright) \end{gathered}`

We then proceed to plot the graph. We have: