What is the equation of the parabola that has a vertex at  (-4, 16)  and passes through point  (0,0) ?

Feb 9, 2016

Let us solve this problem by substituing both points into a parabola equation: $a {x}^{2} + b x + c = y \left(x\right)$

Explanation:

• First of all, let us substitute $\left(0 , 0\right)$:

$a {x}^{2} + b x + c = y \left(x\right) \rightarrow a \cdot {0}^{2} + b \cdot 0 + c = y \left(0\right) \rightarrow c = 0$

Thus, we obtain the independent term in equation, getting $a {x}^{2} + b x = y \left(x\right)$.

• Now, let us substitute the vertex, $\left(- 4 , 16\right)$. We get:

$a \cdot {\left(- 4\right)}^{2} + b \cdot \left(- 4\right) = 16 \rightarrow 16 a - 4 b = 16 \rightarrow 4 a - b = 4$

Now, we have a relation between $a$ and $b$, but we cannot determine them uniquely. We need a third condition.

• For any parabola, the vertex can be obtained by:

${x}_{\text{vertex}} = \frac{- b}{2 a}$

In our case:

${x}_{\text{vertex}} = - 4 = \frac{- b}{2 a} \rightarrow b = 8 a$

• Finally, we must solve the system given by:

{4a-b=4 ; b = 8a}

Replacing $b$ from second equation to the first one:

$4 a - \left(8 a\right) = 4 \rightarrow - 4 a = 4 \rightarrow a = - 1$

And finally:

$b = - 8$

This way, the parabola equation is:

$y \left(x\right) = - {x}^{2} - 8 x$