Question

How do you write the equation of the parabola in vertex form given vertex (4,-1) and y intercept (0,15)?

Answer 1

See below.

Explanation:

Equation in vertex form is `a(x - h)^2 +k` where `h` is the axis of symmetry and `k` is maximum or minimum value.
In this example `h` is `- 4` and `k` is `- 1`:
So
`a(x + 4)^2 - 1` writing as a function: `y = a(x + 4)^2 - 1`

`y` intercept is `( 0 , 15 )`:

Hence: `y = a(x + 4)^2 - 1` => `15 = a( 0 + 4)^2 - 1`
Solving for `a` gives `a = 15/16`

`15/16( x + 4)^2 - 1`

Answer 2

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

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where (h , k ) are the coordinates of the vertex and a is a constant.

`"here " (h,k)=(4,-1)`

`rArry=a(x-4)^2-1`

`"to find a substitute "(0,15)" into the equation"`

`15=16a-1rArra=1`

`rArry=(x-4)^2-1larrcolor(red)" in vertex form"`
graph{(x-4)^2-1 [-10, 10, -5, 5]}