# How do you find the vertex of a parabola y = -2(x+1)(x-5)?

Jun 30, 2015

The vertex is at the point $\left(2 , 18\right)$, that is $x = 2$, $y = 18$

#### Explanation:

Consider a parabola represented by a quadratic polynomial
$y = a {x}^{2} + b x + c$
If $a > 0$ the parabola's horns are directed upwards and it reaches its minimum at some central point $x = {x}_{0}$.
If $a < 0$ the parabola's horns are directed downwards and it reaches its maximum at some central point $x = {x}_{0}$.
In both cases the X-coordinate of a vertex is ${x}_{0}$ and its Y-coordinate is $a {x}_{0}^{2} + b {x}_{0} + c$.

Parabola is a symmetrical figure, its vertical axis of symmetry intersects the X-axis in exactly the same point where it reaches its minimum (for $a > 0$) or maximum (for $a < 0$) value, that is at point $x = {x}_{0}$.

If we are dealing with a quadratic polynomial that can be equal to zero at some values of $x$, that is if the equation
$a {x}^{2} + b x + c = 0$
has solutions $x = {x}_{1}$ and $x = {x}_{2}$,
the central point ${x}_{0}$ lies exactly in the middle between these solutions ${x}_{1}$ and ${x}_{2}$.

Therefore, in case we have two solutions mentioned above, all we have to do to find a vertex is
(a) find these two solutions ${x}_{1}$ and ${x}_{2}$ of an equation:
$a {x}^{2} + b x + c = 0$
(b) find a midpoint between them:
${x}_{0} = \frac{{x}_{1} + {x}_{2}}{2}$.

The given problem presents an easy case, when the solutions to an equation $- 2 \left(x + 1\right) \left(x - 5\right) = 0$ are, obviously, ${x}_{1} = - 1$ and ${x}_{2} = 5$.
Hence, the midpoint between them is
${x}_{0} = \frac{- 1 + 5}{2} = 2$

So, we have determined that X-coordinate of the parabola's axis of symmetry as ${x}_{0} = 2$.
The Y-coordinate is ${y}_{0} _ = - 2 \left(2 + 1\right) \left(2 - 5\right) = 18$.
So the vertex is at point $\left(2 , 18\right)$