Question

Can i have some help rn?

Answer 1

• Vertex at (-1,-4)
• Passes through (1,0)

Find a

• a(x-h)²+k=y
• a(1+1)²-4=0
• 4a-4=0
• 4a=4
• a=1

So

Vertex form

• y=(x+1)²-4

#2

• (h,k)=(4,-2)
• passes through (6,6)

Find a

• a(6-4)²-2=6
• 4a=8
• a=2

Vertex form

• y=2(x-4)²-2

Answer 2

`\textsf{a)} \quad y=(x+1)^2-4`

`\textsf{b)} \quad y=2(x-4)^2-2`

Explanation:

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

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

Part a

• Vertex = (-1, -4)
• Point on parabola = (1, 0)

Substitute the vertex into the formula:

`\begin{aligned}\implies y &=a(x-(-1))^2-4\\y & =a(x+1)^2-4\end{aligned}`

Substitute the point (1, 0) into the formula:

`\begin{aligned}\implies a(1+1)^2-4&=0\\4a-4&=0\\4a &=4\\a &=1\end{aligned}`

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

`y=(x+1)^2-4`

Part b

• Vertex = (4, -2)
• Point on parabola = (6, 6)

Substitute the vertex into the formula:

`\begin{aligned}\implies y &=a(x-4)^2+(-2)\\y & =a(x-4)^2-2\end{aligned}`

Substitute the point (6, 6) into the formula:

`\begin{aligned}\implies a(6-4)^2-2&=6\\4a-2 & =6\\4a &=8\\a &=2\end{aligned}`

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

`y=2(x-4)^2-2`