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

Aug 7, 2017

Axis of symmetry = 2
Vertex = (2,-4)
X-intercepts = 0,4

#### Explanation:

The axis a symmetry can be found using a simple formula of

$x = - \frac{b}{2 a}$

when the graph is in the form of

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

In your graph, $a$ would equal 1 and $b$ would equal -4. Substituting this into the equation would give you the axis of symmetry of

x=(-(-4))/(2(1)

$x = 2$

Additionally, the vertex also resides on the axis of symmetry. We just need to find the y-coordinate of the vertex, which we can do by substituting $x = 2$ back into the orginal equation giving us

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

$y = - 4$

Thus, the vertex will lie on $\left(2 , - 4\right)$

To find the x-intercepts, we must find where the graph intersects the x-axis and this can be done when we let the graph equal zero.

$0 = {x}^{2} - 4 x$

Factorising we get

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

Finally, using the null factor law, we can find what the x-intercepts are.

$x = 0 , 4$

Aug 7, 2017

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

#### Explanation:

$y = {x}^{2} - 4 x = {x}^{2} - 4 x + 4 - 4 = {\left(x - 2\right)}^{2} - 4$ Comparing with

Standard vertex fom of equation  y = a (x-h)^2 ; (h,k)  being

vertex we find here $h = 2 , k = - 4$, so vertex is at $\left(2 , - 4\right)$.

Axis of symmetry is x = 2 ; x  intercepts can be found by putting

$y = 0$ in the equation .

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

$\therefore x = 0 \mathmr{and} x = 4$ So $x$ intercepts are $x = 0 , x = 4$

Vertex is at $\left(2 , - 4\right)$ , axis of symmetry is $x = 2$ and

$x$ intercepts are $x = 0 , x = 4$.

graph{x^2-4x [-10, 10, -5, 5]} [Ans]