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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