# How do you find its vertex, axis of symmetry, y-intercept and x-intercept for f(x) = x^2 - 4x?

Apr 21, 2018

Vertex is $\left(2 , - 4\right)$; axis of symmetry is $x - 2 = 0$; $x$-intercepts are $0$ and $4$ and $y$-intercept is $0$.

#### Explanation:

This is the equation of parabola. Intercept form of equation of parabola is $f \left(x\right) = a \left(x - \alpha\right) \left(x - \beta\right)$, where $\alpha$ and $\beta$ are intercepts on $x$-axis.

As we have $f \left(x\right) = {x}^{2} - 4 x = x \left(x - 4\right) = \left(x - 0\right) \left(x - 4\right)$, $x$-intercepts are $0$ and $4$.

$y$-intercept is obtained by putting $x = 0$ and hence it is ${0}^{2} - 4 \cdot 0 = 0 - 0 = 0$ and hence $y$-intercept is $0$.

Vertex form of equation of parabola is $f \left(x\right) = a {\left(x - h\right)}^{2} + k$, where $\left(h , k\right)$ is vertex and $x - h = 0$ is axis of symmetry

Here we have $f \left(x\right) = {x}^{2} - 4 x = \left({x}^{2} - 4 x + 4\right) - 4 = {\left(x - 2\right)}^{2} - 4$

Hence axis of symmetry is $x - 2 = 0$ and vertex is $\left(2 , - 4\right)$.

graph{(x^2-4x-y)(x-2)((x-2)^2+(y+4)^2-0.03)=0 [-9.92, 10.08, -5.12, 4.88]}