Question

Could someone help me rnnn?

Answer 1

It can be solved through some shortcut tricks of parabola

• y intercept= (0,-4)

It passes through (-2,8) and (1,-1)

The minimum equation of the parabola (as it's quadratic) for vertex at (0,0)

• y=ax²

whatever our required parabola be it's translated from the above one

So

• ax²=y

Let a be 0

• current equation is y=x²

So it has vertex as well as y inetrcept at (0,0)

For x=-2

• y=(-2)²=4 (but its 8 in our translated one )

For x=1

• y=1

It wasn't negative but in translation it's negative so something is subtracted

Take a =2 and subtract (-2)² i.e 4 as it's minimal value of y=x² where x≠y

For

(-2,8)

• -2=2(4)-4=8-4=4

No try a=3

• -2=3(4)-4=12-4=8

Yes satisfied

Take (1,-1)

• -1=3(1)-4=3-4=-1

Verified

The required equation is

• y=3x²-4

Answer 2

vertex = (0, -4)

equation of the parabola:
`y=3x^2-4`

Explanation:

Given:

• y-intercept of parabola: -4
• parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola:
`y=a(x-h)^2+k`

(where (h, k) is the vertex and
`a` is some constant)

Substitute point (0, -4) into the equation:

`\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}`

Substitute point (-2, 8) and
`ah^2+k=-4` into the equation:

`\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}`

Substitute point (1, -1) and
`ah^2+k=-4` into the equation:

`\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}`

Equate to find h:

`\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}`

Substitute found value of h into one of the equations to find a:

`\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}`

Substitute found values of h and a to find k:

`\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}`

Therefore, the equation of the parabola in vertex form is:

`\implies y=3(x-0)^2-4=3x^2-4`

So the vertex of the parabola is (0, -4)