# How do you find the vertex of the parabola y = x^2 - 4x?

May 20, 2018

$\left(2 , - 4\right)$

#### Explanation:

The easiest way is:

$y = a {x}^{2} + b x + c$

axis on symmetry is:

$a o s = \frac{- b}{2 a}$

Vertex is: $\left(a o s , f \left(a o s\right)\right)$

c = y-intercept

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

a = 1
b = -4
c = 0

$a o s = \frac{- \left(- 4\right)}{2 \cdot 1} = 2$

f(aos) means we put the aos back in your function as x and solve for y:

$f \left(a o s\right) = f \left(2\right) = {2}^{2} - 4 \cdot 2 = - 4$

Vertex is: $\left(a o s , f \left(a o s\right)\right)$

Vertex is: $\left(2 , - 4\right)$

Note, this can also be solved by completing the square and converting the function to vertex form.

graph{y = x^2 - 4x [-7.13, 12.87, -7.8, 2.2]}

May 20, 2018

$\text{vertex } = \left(2 , - 4\right)$

#### Explanation:

$\text{the vertex lies on the axis of symmetry which is}$
$\text{situated at the midpoint of the zeros}$

$\text{to find the zeros set y = 0}$

$\Rightarrow {x}^{2} - 4 x = 0 \leftarrow \textcolor{b l u e}{\text{factorise}}$

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

$\text{equate each factor to zero and solve for x}$

$\Rightarrow x = 0$

$x - 4 = 0 \Rightarrow x = 4$

$\text{axis of symmetry/x-coordinate of vertex } = \frac{0 + 4}{2} = 2$

$\text{substitute this value into the equation for y}$

$\Rightarrow y = {2}^{2} - \left(4 \times 2\right) = 4 - 8 = - 4$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(2 , - 4\right)$
graph{(y-x^2+4x)((x-2)^2+(y+4)^2-0.04)=0 [-10, 10, -5, 5]}