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

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

#x=-b/(2a)#

when the graph is in the form of

#y=ax^2+bx+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=(2)^2#-4(2)

#y=-4#

Thus, the vertex will lie on #(2,-4)#

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^2-4x#

Factorising we get

#0=x(x-4)#

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

#x=0,4#

# y = x^2 -4x = x^2 -4x +4 - 4 = (x-2)^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 #(2,-4)#.

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

#y=0# in the equation .

# y= x^2-4x or y= x(x-4) or x(x-4)=0 #

#:. x=0 or x=4 # So #x # intercepts are #x=0 ,x=4#

Vertex is at #(2,-4)# , axis of symmetry is #x = 2# and

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

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