# How do you find two quadratic function one that opens up and one that opens downward whose graphs have intercepts (4,0), (8,0)?

Jan 29, 2017

$y = {x}^{2} - 12 x + 32 \leftarrow \text{ upwards}$
$y = - {x}^{2} + 12 x - 32 \leftarrow \text{ downwards}$

#### Explanation:

There are an infinite number of quadratic functions possible.

Since we know the x-intercepts (4 ,0) and (8 ,0)

$\Rightarrow \text{roots are " x=4" and } x = 8$

$\text{and the factors are } \left(x - 4\right) , \left(x - 8\right)$

Hence the general equation is.

$y = a \left(x - 4\right) \left(x - 8\right)$

• If a > 0 , then graph opens up

• If a < 0 , then graph opens down

To find the particular function we require to know another point

For this example, lets choose a = 1 and a = - 1

$a = 1 \to y = {x}^{2} - 12 x + 32 \leftarrow \text{ distributing brackets}$

$a = - 1 \to y = - \left({x}^{2} - 12 x + 32\right) = - {x}^{2} + 12 x - 32$
graph{(y-x^2+12x-32)(y+x^2-12x+32)=0 [-10, 10, -5, 5]}