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:

$Vertex,(h,k)=\mleft(-\frac{b}{2a},\frac{4ac-b^2}{4a}\mright)$

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

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

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: