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

1 Answer
Shwetank Mauria
Apr 21, 2018

Vertex is #(2,-4)#; 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(x)=a(x-alpha)(x-beta)#, where #alpha# and #beta# are intercepts on #x#-axis.

As we have #f(x)=x^2-4x=x(x-4)=(x-0)(x-4)#, #x#-intercepts are #0# and #4#.

#y#-intercept is obtained by putting #x=0# and hence it is #0^2-4*0=0-0=0# and hence #y#-intercept is #0#.

Vertex form of equation of parabola is #f(x)=a(x-h)^2+k#, where #(h,k)# is vertex and #x-h=0# is axis of symmetry

Here we have #f(x)=x^2-4x=(x^2-4x+4)-4=(x-2)^2-4#

Hence axis of symmetry is #x-2=0# and vertex is #(2,-4)#.

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

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