Question

Can someone help me rn?

Answer 1

• y=a(x-1)²+q

Vertex at (1,q)

There is a point already given from which parabola passes through its (1,1)

• So q=1

Hence

• vertex=(1,1)

Equation:-

• y=a(x-h)²+k
• y=a(x-1)²+1

Now put (-1,-9)

• -9=a(-2)²+1
• -9=4a+1
• 4a=-10
• a=-5/2

a is negative hence parabola is opening downwards

So

vertex is maximum

Max value

• y=1

Answer 2

y = 1

Explanation:

Given:

• equation of the parabola:
  `y=a(x-1)^2+q`
• points on the parabola: (-1, -9) and (1, 1)

Substitute point (1, 1) into the given equation of the parabola to find q:

`\begin{aligned}\textsf{At}\:(1,1)\implies a(1-1)^2+q &=1\\a(0)^2+q &=1\\q &=1\end{aligned}`

Substitute the found value of q and point (-1, -9) into the given equation of the parabola to find a:

`\begin{aligned}\textsf{At}\:(-1,-9)\implies a(-1-1)^2+1 &=-9\\a(-2)^2+1 &=-9\\4a+1 &=-9\\4a &=-10\\a &=-(5)/(2)\end{aligned}`

Therefore, the equation of the parabola is:

`y=-(5)/(2)(x-1)^2+1`

The maximum or minimum point on a parabola is the vertex.

Vertex form of a parabola:
`y=a(x-h)^2+k` where (h, k) is the vertex.

Therefore, the vertex of the parabola is (1, 1) and so the maximum value of y = 1