# What is the equation of the parabola that has a vertex at  (4, 6)  and passes through point  (-7,-8) ?

Mar 31, 2016

$y = \left(- \frac{14}{121}\right) {x}^{2} + \frac{112}{121} x + \frac{502}{121}$

$\textcolor{g r e e n}{\text{Some one may be able to demonstrate a simpler method.}}$

#### Explanation:

Known: The equation standard form is $y = a {x}^{2} + b x + c$

Using the two points to give simultaneous equations

$- 8 = a {\left(- 7\right)}^{2} + b \left(- 7\right) + c$.................................(1)
$\text{ } 6 = a {\left(4\right)}^{2} + b \left(4\right) + c$............................................(2)

Known that ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \left(\frac{b}{a}\right)$

$\implies 4 = \left(- \frac{1}{2}\right) \left(\frac{b}{a}\right)$

So $a \left(2 \times 4\right) = - b$

$\implies b = - 8 a$....................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Substitute (3) into (1) and (2)}}$

$- 8 = a {\left(- 7\right)}^{2} + \left(- 8 a\right) \left(- 7\right) + c$
$\text{ } 6 = a {\left(4\right)}^{2} + \left(- 8 a\right) \left(4\right) + c$

$- 8 = 49 a + 56 a + c$
$\text{ } 6 = 16 a - 32 a + c$

$- 8 = \text{ "105a+c" } \ldots \ldots \ldots \ldots \ldots \ldots . \left({1}_{a}\right)$
$\text{ "6=-16a+c" } \ldots \ldots \ldots \ldots \ldots . \left({2}_{a}\right)$

$\textcolor{b r o w n}{\text{Subtract "(2_a)" from } \left({1}_{a}\right)}$

$- 14 = 121 a$

$\textcolor{b l u e}{a = - \frac{14}{121}}$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\implies b = - 8 a \to b = \left(- 8\right) \times \left(- \frac{14}{121}\right) = + \frac{112}{121}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b r o w n}{\text{Substitute for "a" in } \left({2}_{a}\right)}$

$\text{ } 6 = - 16 \left(- \frac{14}{121}\right) + c$

$\textcolor{b l u e}{c = 4 \frac{18}{121} = \frac{502}{121}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mar 31, 2016

In vertex form:

$y = - \frac{14}{121} {\left(x - 4\right)}^{2} + 6$

In standard polynomial form:

$y = - \frac{14}{121} {x}^{2} + \frac{112}{121} x + \frac{502}{121}$

#### Explanation:

The equation of a parabola with vertical axis can be expressed in vertex form as:

$y = a {\left(x - h\right)}^{2} + k$

where $\left(h , k\right)$ is the vertex and $a$ a multiplier.

In our example, the equation of the parabola can be written in the form:

$y = a {\left(x - 4\right)}^{2} + 6$

from which we can deduce:

$a = \frac{y - 6}{x - 4} ^ 2$

Since we want our parabola to pass through $\left(- 7 , - 8\right)$ substitute these values of $x$ and $y$ into this equation to find:

$a = \frac{- 8 - 6}{- 7 - 4} ^ 2 = - \frac{14}{121}$

So the equation of our parabola may be written:

$y = - \frac{14}{121} {\left(x - 4\right)}^{2} + 6$

$= - \frac{14}{121} {x}^{2} + \frac{112}{121} x - \frac{224}{121} + 6$

$= - \frac{14}{121} {x}^{2} + \frac{112}{121} x + \frac{502}{121}$

graph{-14/121 x^2 + 112/121 x +502/121 [-20, 20, -10, 10]}