Question

Plot five points on the parabolic,the vertex, two points to the left of the vertex, and two points to the right of the vertex

Answer 1

Explanation

Step 1

find the vertex,

when you have a equation in the standard form

$y=ax^2+bx+c$

the vertex is given by

$x=-\frac{b}{2a}$

hence,

$\begin{gathered} y=ax^2+bx+c\Rightarrow y=-2x^2 \\ so \\ a=-2 \\ b=0 \\ so,\text{ x= -}\frac{0}{2(-2)}=0 \\ so,\text{ the vertex is x= 0 ( or y- A}\Xi s) \end{gathered}$

evaluate when x= 0 to find the coordinate of the vertex

$\begin{gathered} y=-2(0)^2=0 \\ so,\text{ vertex=(0,0)} \end{gathered}$

P1(0,0)

Step 2

now, 2 poins to the left

a)

for x=-1

$\begin{gathered} y=-2x^2 \\ y=-2(-1)^2=-2(1)=-2 \\ \text{hence} \\ P2(-1,-2) \end{gathered}$

P2(-1,-2)

b)when x=-2

$\begin{gathered} y=-2x^2 \\ y=-2(-2)^2=-2(4)=-8 \\ \text{hence} \\ P3(-2,-8) \end{gathered}$

Step 3

2 points to the rigth

a)

for x=1

$\begin{gathered} y=-2x^2 \\ y=-2(1)^2=-2(1)=-2 \\ \text{hence} \\ P4(1,-2) \end{gathered}$

P4(1,-2)

b)when x=2

$\begin{gathered} y=-2x^2 \\ y=-2(2)^2=-2(4)=-8 \\ \text{hence} \\ P3(2,-8) \end{gathered}$

Step 4

I hope this helps you