# How do you find two quadratic function one that opens up and one that opens downward whose graphs have intercepts (0,0), (10,0)?

Jan 13, 2017

${f}_{\text{up}} \left(x\right) = {x}^{2} - 10 x$

${f}_{\text{down}} \left(x\right) = - {x}^{2} + 10 x$

#### Explanation:

Any quadratic function with these $x$ intercepts is expressible in the form:

$f \left(x\right) = a x \left(x - 10\right) = a {x}^{2} - 10 a x \text{ }$ for some constant $a \ne 0$

It must take this form in order that $f \left(0\right) = 0$ and $f \left(10\right) = 0$.

If $a > 0$ then this will open upwards and if $a < 0$ then it will open downwards.

So we can use $a = \pm 1$ to define the functions:

${f}_{\text{up}} \left(x\right) = {x}^{2} - 10 x$

${f}_{\text{down}} \left(x\right) = - {x}^{2} + 10 x$