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

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Answer 1 · 2015-12-19T16:39:21

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

The parabola $f$ $f$ is written as $a x^{2} + b x + c$ $ax^2 + bx + c$ such that $a \neq 0$ $a != 0$ .

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

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

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 $a = 3/16$ .

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

$6 = -9a + c iff c = 6 + 9*(3/16) iff c = 123/16$ and $b = 6a iff b = 9/8$ .

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