Question

I need help finding 5 points. the vertex, 2 to the left of vrrtex, and 2 to the right of vertex

Answer 1

Given the equation of the parabola:

`y=-3x^2-12x-7`

Step 1: Determine the vertex of the parabola.

First, find the vertex using the vertex formula below:

`Vertex,(h,k)=\mleft(-(b)/(2a),(4ac-b^2)/(4a)\mright)`

From the equation: a=-3, b=-12, and c=-7

`\begin{gathered} Vertex,(h,k)=\mleft(-(-12)/(2*-3),(4(-3)(-7)-(-12)^2)/(4*-3)\mright) \\ =(-2,(-60)/(-12)) \\ =(-2,5) \end{gathered}`

Step 2: Find two points to the left of the vertex.

When x=-3

`\begin{gathered} y=-3(-3)^2-12(-3)-7 \\ =-3(9)+36-7 \\ =2 \\ \implies(-3,2) \end{gathered}`

When x=-4

`\begin{gathered} y=-3(-4)^2-12(-4)-7 \\ =-3(16)+48-7 \\ =-7 \\ \implies(-4,-7) \end{gathered}`

The two points to the left are (-3, 2) and (-4, -7).

Step 3: Find two points to the right of the vertex.

When x=-1

`\begin{gathered} y=-3(-1)^2-12(-1)-7 \\ =-3(1)+12-7 \\ =2 \\ \implies(-1,2) \end{gathered}`

When x=0

`\begin{gathered} y=-3(0)^2-12(0)-7 \\ =0+0-7 \\ =-7 \\ \implies(0,-7) \end{gathered}`

The two points to the left are (-1, 2) and (0, -7).

The graph showing the 5 points is given below: