# What is the the vertex of y = 1/2(x+1)(x-5) ?

Dec 13, 2015

$y = \frac{1}{2} {\left(x - \textcolor{red}{2}\right)}^{2} \textcolor{b l u e}{- \frac{9}{2}}$

vertex: $\left(2 , - \frac{9}{2}\right)$

#### Explanation:

Note:

Vertex form $f \left(x\right) = a {\left(x - h\right)}^{2} + k$
$h = {x}_{v e r t e x} = - \frac{b}{2 a} \text{ " " }$ ; $k = {y}_{v e r t e x} = f \left(- \frac{b}{2 a}\right)$

Given:

$y = \frac{1}{2} \left(x + 1\right) \left(x - 5\right)$

Multiply the expression or FOIL

$y = \frac{1}{2} \left({x}^{2} - 5 x + x - 5\right)$

$y = \frac{1}{2} \left({x}^{2} - 4 x - 5\right)$

$y = \frac{1}{2} {x}^{2} - 2 x - \frac{5}{2}$

a = 1/2;" " b= -2;" " " c= -5/2

$\textcolor{red}{h = {x}_{v e r t e x}} = \frac{- \left(- 2\right)}{2 \cdot \frac{1}{2}} = \textcolor{red}{2}$
$\textcolor{b l u e}{k = {y}_{v e r t e x}} = f \left(2\right) = \frac{1}{2} {\left(2\right)}^{2} - 2 \left(2\right) - \frac{5}{2}$

=> 2-4 -5/2 => -2 -5/2 => color(blue)(-9/2

The vertex form is

$y = \frac{1}{2} {\left(x - \textcolor{red}{2}\right)}^{2} \textcolor{b l u e}{- \frac{9}{2}}$

Dec 13, 2015

$\left(2 , - \frac{9}{2}\right)$

#### Explanation:

First, find the expanded form of the quadratic.

$y = \frac{1}{2} \left({x}^{2} - 4 x - 5\right)$

$y = \frac{1}{2} {x}^{2} - 2 x - \frac{5}{2}$

Now, the vertex of a parabola can be found with the vertex formula:

$\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$

Where the form of a parabola is $a {x}^{2} + b c + c$.

Thus, $a = \frac{1}{2}$ and $b = - 2$.

The $x$-coordinate is $- \frac{- 2}{2 \left(\frac{1}{2}\right)} = 2$.

The $y$-coordinate is $f \left(2\right) = \frac{1}{2} \left(2 + 1\right) \left(2 - 5\right) = - \frac{9}{2}$

Thus, the vertex of the parabola is $\left(2 , - \frac{9}{2}\right)$.

You can check the graph:
graph{1/2(x+1)(x-5) [-10, 10, -6, 5]}

Dec 14, 2015

$\textcolor{b l u e}{\text{A slightly quicker approach}}$

$\textcolor{g r e e n}{\text{It is not uncommon for there to be several ways of solving a problem!}}$

#### Explanation:

This is a quadratic thus of the hors shoe type shape.

That means that the vertex is $\frac{1}{2}$ way between the x-intercepts.

The x-intercepts will occur when y=0

If y is 0 then the right side also = 0

The right side equals zero when $\left(x + 1\right) = 0 \text{ or } \left(x - 5\right) = 0$

For $\left(x + 1\right) = 0 \to x = - 1$
For$\left(x - 5\right) = 0 \to x = + 5$

Half way is $\frac{\left(- 1\right) + \left(+ 5\right)}{2} = \frac{4}{2} = 2$

Having found $\textcolor{b l u e}{{x}_{\text{vertex}} = 2}$ we then substitute in the original equation to find $\textcolor{b l u e}{{y}_{\text{vertex}}}$