# 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)?

Jun 7, 2015

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

Every quadratic function can be expressed in the standard form

$f \left(x\right) = a {x}^{2} + b x + c$

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

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$, where $\left(h , k\right)$ is the vertex.

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

$f \left(x\right) = a {\left(x - 2\right)}^{2} + 9$

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

$f \left(1\right) = a {\left(1 - 2\right)}^{2} + 9 = 0$. This gives $a = - 9$
$f \left(3\right) = a {\left(3 - 2\right)}^{2} + 9 = 0$. This also gives $a = - 9$

Therefore $f \left(x\right) = - 9 {\left(x - 2\right)}^{2} + 9$. Factor it out and simplify to get the standard form:

$f \left(x\right) = - 9 {x}^{2} + 36 x - 27$