Question

How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (1,0) & (3,0)?

Answer 1

You've got to know the vertex form of a quadratic function!

Every quadratic function can be expressed in the standard form

`f(x)=ax^2+bx+c`

Using factoring and completing the square, many Algebra textbooks will give you a rewritten form of the above equation:

`f(x)= a(x-h)^2+k`, where `(h,k)` is the vertex.

Because `h=2` and `k=9`, we can plug these into the rewritten `f(x)` to get:

`f(x)=a(x-2)^2+9`

The roots let us solve for `a`. The first root says `f(1)=0` and the second root says `f(3)=0`. Therefore,

`f(1) = a(1-2)^2+9=0`. This gives `a=-9`
`f(3) = a(3-2)^2+9=0`. This also gives `a=-9`

Therefore `f(x)=-9(x-2)^2+9`. Factor it out and simplify to get the standard form:

`f(x) = -9x^2+36x-27`