# How do you find the vertex and intercepts for y = x^2 - 4x?

Jul 6, 2017

See below.

#### Explanation:

$x$-intercepts:
Solve ${x}^{2} - 4 x = 0$

$x \left(x - 4\right) = 0$,

$x = 0 , 4$

$y$-intercept:
When $x = 0$, the equation gives us $y = 0$

Vertex: is midway between the intercepts. The midpoint is the average.

$x = \frac{0 + 4}{2} = 2$

When $x = 2$, the equation gives us $y = {\left(2\right)}^{2} - 4 \left(2\right) = 4 - 8 = - 4$

The vertex is $\left(2 , - 4\right)$

Jul 6, 2017

A different method that uses $\textcolor{red}{\text{part of}}$ completing the square. Purely given to show that sometimes different methods work.

#### Explanation:

Given: $y = {x}^{2} - 4 x$

$\textcolor{b l u e}{\text{Determine y-intercept}}$

Write as $y = {x}^{2} - 4 x + 0$
$\text{ } \textcolor{red}{\uparrow}$
" "color(red)("y-intercept")

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine vertex}}$

Note that in the standardised form of $y = a {x}^{2} + b x + c$ our equation is such that $a = 1$ so $\frac{b}{a} = \frac{- 4}{1} = - 4$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a} \to \left(- \frac{1}{2}\right) \times \left(- 4\right) = + 2$

By substitution

y_("vertex")=(2)^2-4(2)" "=" "4-8" "=" "-8

Vertex$\to \left(x , y\right) = \left(2 , - 8\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine x-intercept}}$

The ${x}^{2}$ term is positive so the graph is of form $\cup$ thus has a minimum. ${y}_{\text{vertex}} < 0$ so x-intercepts exist.

Factorising

The common term is $x$ so lets factor that out giving.

$y = x \left(x - 4\right)$

The x-intercepts are at $y = 0$ so by substitution

$y = 0 = {x}^{2} \left(x - 4\right)$

Thus $x = 0 \mathmr{and} 4$