Question

I need help with 5 points please. the vertex. 2 to the left of vertex. 2 to the right of vertex.

Answer 1

Step 1

Given;

`y=-x^2+10x-24`

Required; To graph the function.

Step 2

Graph the parabola showing the vertex with two points to the left and the right of the vertex respectively.

`\begin{gathered} \text{For the vertex(h,k); } \\ h=-(b)/(2a) \\ b=10 \\ a=-1,c=-24 \\ h=-(10)/(2(-1)) \\ h=-(10)/(-2) \\ h=5 \\ k=(-D)/(4a) \\ D=b^2-4ac \end{gathered}`
`\begin{gathered} D=(10)^2-(4(-1)(-24)) \\ D=100-96 \\ D=4 \\ k=(-4)/(4(-1))=(-4)/(-4)=1 \\ \text{The vertex, (h,k)=(5,1)} \end{gathered}`

Step 3

Two points to the right

`\begin{gathered} when\text{ x=6} \\ y=-(6)^2+10(6)-24 \\ y=-36+60-24 \\ y=0 \\ when\text{ x=8} \\ y=-(8)^2+10(8)-24 \\ y=-64+80-24 \\ y=-8 \\ \text{The two points are;} \\ 1)_{}(6,0) \\ 2)(8,-8) \end{gathered}`

Step 4

Two points the left

`\begin{gathered} When\text{ x=4} \\ y=-(4)^2+10(4)-24 \\ y=-16+40-24=0 \\ \text{when x=}2 \\ y=-(2)^2+10(2)-24 \\ y=-4+20-24=-8 \\ \text{The two point are;} \\ 1)(4,0) \\ 2)(2,-8) \end{gathered}`

Step 5

Plot the graph