# What is the equation of the parabola that has a vertex at  (-3, 6)  and passes through point  (1,9) ?

Dec 19, 2015

$f \left(x\right) = \frac{3}{16} {x}^{2} + \frac{9}{8} x + \frac{123}{16}$

#### Explanation:

The parabola $f$ is written as $a {x}^{2} + b x + c$ such that $a \ne 0$.

1st of all, we know this parabol has a vertex at $x = - 3$ so $f ' \left(- 3\right) = 0$. It already gives us $b$ in function of $a$.

$f ' \left(x\right) = 2 a x + b$ so $f ' \left(- 3\right) = 0 \iff - 6 a + b = 0 \iff b = 6 a$

We now have to deal with two unknown parameters, $a$ and $c$. In order to find them, we need to solve the following linear system :

6 = 9a - 18a + c; 9 = a + 6a + c iff 6 = -9a + c;9 = 7a + c

We now substract the 1st line to the 2nd one in the 2nd line :

6 = -9a + c;3 = 16a so we now know that $a = \frac{3}{16}$.

We replace $a$ by its value in the 1st equation :

$6 = - 9 a + c \iff c = 6 + 9 \cdot \left(\frac{3}{16}\right) \iff c = \frac{123}{16}$ and $b = 6 a \iff b = \frac{9}{8}$.