# What is the equation of the parabola that has a vertex at  (3, -5)  and passes through point  (13,43) ?

Aug 29, 2016

$\textcolor{b l u e}{\text{I have taken you to a point from which you can take over}}$

#### Explanation:

Let the point ${P}_{1} \to \left(x , y\right) = \left(13 , 43\right)$

Quadratic standard form equation: $y = a {x}^{2} + b x + 5 \textcolor{w h i t e}{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q n \left(1\right)$

Vertex form equation: $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k \textcolor{w h i t e}{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q n \left(2\right)$

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$\textcolor{b r o w n}{\text{Using Eqn(2)}}$

We are given that Vertex$\to \left({x}_{\text{vertex"),y_("vertex}}\right) = \left(3 , - 5\right)$

But ${x}_{\text{vertex")=(-1)xxb/(2a)=+3" "=>" "b=-6acolor(white)(" }} \ldots \ldots E q n \left(3\right)$

Side note: $k = - 5$ from vertex y-coordinate
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$\textcolor{b r o w n}{\text{Using Eqn(3) substitute for b in Eqn(1)}}$

$y = a {x}^{2} + \left(- 6 a\right) x + 5$ ...........................Eqn(4)

But we are given the point ${P}_{1} \to \left(13 , 43\right)$

Thus Eqn(4) becomes:

$43 = a {\left(13\right)}^{2} - 6 a \left(13\right) + 5 \textcolor{w h i t e}{\text{ }} \ldots \ldots E q n \left({4}_{a}\right)$

$\textcolor{b l u e}{\text{From this you can solve for "a" and from that solve for } b}$

$\textcolor{red}{\text{I will let you take over from this point}}$