Question

How do you write a quadratic equation with Vertex- (5,5) Point- (6,6)?

Answer 1

`y=x^2-10x+30`

Explanation:

Two solve the question the two forms of a parabola's equation should be known:
Standard form: `y=ax^2+bx+c`
Vertex form `y=a(x-h)^2+k` (with vertex at (`h,k`))

First we input values from the question into the vertex form equation:
`y=a(x-h)^2+k`
`=a(x-5)^2+5`
`=a(x-5)^2+5`

Still there is the unknown `a` for which we need to find its value. Using the other piece of information, we can find `a`:
`6=a(6-5)^2+5`
`6=a+5`
`a=1`

Now that we know `a`, we can write an equation with no unknowns and convert it to its standard form:
`y=(x-5)^2+5`
`y=(x^2-10x+25)+5`
`y=x^2-10x+25+55`
`y=x^2-10x+30`

Answer 2

Using the standard vertex form for a parabola with a vertical axis of symmetry:
`color(white)("XXX")y=m(x-a)^2+b` with its vertex at `(a,b)`

A parabola with a vertical axis of symmetry and a vertex at `(5,5)`
would have the form:
`color(white)("XXX")y=m(x-5)^2+5` for some constant value `m`

If `(x,y)=(6,6)` is a point on the required equation, then
`color(white)("XXX")6=m(6-5)^2+5`
`color(white)("XXX")m=1`
and the equation would be:
`color(white)("XXX")y=1(x-5)^2+5`

This could be rearranged as
`color(white)("XXX")y=x^2-10x+30` (standard quadratic form)#

For a parabola with a horizontal axis of symmetry,
the `x` and `y` values could be exchanged giving
`color(white)("XXX")x=y^2-10y+30`
as an alternative solution that meets the given requirements.