Question

What is the equation of the parabola that has a vertex at `(-1, 16)` and passes through point `(3,20)`?

Answer 1

`f(x) = 1/4(x+1)^2 + 16`

Explanation:

The standard form of the equation of a parabola is:
`f(x) = a(x-h)^2 + k`

From the question we know two things.

1. The parabola has a vertex at `(-1, 16)`
2. The parabola passes through the point `(3, 20)`

With those two pieces of information, we can construct our equation for the parabola.

Let's start off with the basic equation:
`f(x) = a(x-h)^2 + k`

Now we can substitute our vertex coordinates for `h` and `k`

The `x` value of your vertex is `h` and the `y` value of your vertex is `k`:

`f(x) = a(x+1)^2 + 16`

Note that putting `-1` in for `h` makes it `(x-(-1))` which is the same as `(x+1)`

Now substitute the point the parabola passes through for `x` and `y` (or `f(x)`):

`20 = a(x+1)^2 + 16`

Looks good. Now we have to find `a`

Combine all like terms:

Add 3 + 1 inside the parentheses:
`20 = a(4)^2 + 16`

Square 4:
`20 = 16a+ 16`

Factor out 16:
`20 = 16(a+1)`

Divide both sides by 16:
`20/16 = a+1`

Simplify `20/16`:
`5/4 = a+1`

Subtract 1 from both sides:
`5/4 -1 = a`

The LCD of 4 and 1 is 4 so `1 = 4/4`:
`5/4 -4/4 = a`

Subtract:
`1/4 = a`

Switch sides if you want:
`a = 1/4`

Now that you've found `a`, you can plug it into the equation with the vertex coordinates:

`f(x) = 1/4(x+1)^2 + 16`

And that's your equation.
Hope this helped.

Answer 2

`y=1/4(x+1)^2+16`

Explanation:

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

`color(red)(bar(ul(|color(white)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 multiplier"`

`"here "(h,k)=(-1,16)`

`rArry=a(x+1)^2+16`

`"to find a substitute "(3,20)" into the equation"`

`20=16a+16rArra=1/4`

`rArry=1/4(x+1)^2+16larrcolor(red)"in vertex form"`