# How do you write a quadratic equation with x-intercepts: -3,2 ; point: (3,6)?

Jul 17, 2016

Use a couple of quadratic properties and algebra to find the equation is $y = {x}^{2} + x - 6$.

#### Explanation:

If a quadratic equation has solutions $x = a$ and $x = b$, then $x - a = 0$ and $x - b = 0$. Furthermore, the quadratic can be written as $y = c \left(x - a\right) \left(x - b\right)$, where $c$ is some constant. The reasoning is that if you set $y$ equal to $0$, you get:
$c \left(x - a\right) \left(x - b\right) = 0$
Which is the same as:
$\left(x - a\right) \left(x - b\right) = 0$
And so the solutions are $x = a$ and $x = b$ - which is exactly what we started with.

Alright, enough theory - let's get on with it! We are told that the $x$-intercepts are $- 3$ and $2$, and since $x$-intercepts are the same thing as zeros, $x = - 3$ and $x = 2$ are solutions. Following the process from above, we can write the quadratic as:
$y = c \left(x + 3\right) \left(x - 2\right)$

To solve for $c$, we use the other piece of info we were given: the point $\left(3 , 6\right)$:
$y = c \left(x + 3\right) \left(x - 2\right)$
$\to 6 = c \left(3 + 3\right) \left(3 - 2\right)$
$\to 6 = c \left(6\right) \left(1\right)$
$\to 6 = 6 c \to c = 1$

So the equation of the quadratic is:
$y = 1 \left(x + 3\right) \left(x - 2\right)$
$\to y = \left(x + 3\right) \left(x - 2\right) = {x}^{2} + 3 x - 2 x - 6 = {x}^{2} + x - 6$