How do you find its vertex, axis of symmetry, y-intercept and x-intercept for f(x) = x^2 - 4xf(x)=x2−4x?

1 Answer
Shwetank Mauria
Apr 21, 2018

Vertex is (2,-4)(2,−4); axis of symmetry is x-2=0x−2=0; xx-intercepts are 00 and 44 and yy-intercept is 00.

Explanation:

This is the equation of parabola. Intercept form of equation of parabola is f(x)=a(x-alpha)(x-beta)f(x)=a(x−α)(x−β), where alphaα and betaβ are intercepts on xx-axis.

As we have f(x)=x^2-4x=x(x-4)=(x-0)(x-4)f(x)=x2−4x=x(x−4)=(x−0)(x−4), xx-intercepts are 00 and 44.

yy-intercept is obtained by putting x=0x=0 and hence it is 0^2-4*0=0-0=002−4⋅0=0−0=0 and hence yy-intercept is 00.

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

Here we have f(x)=x^2-4x=(x^2-4x+4)-4=(x-2)^2-4f(x)=x2−4x=(x2−4x+4)−4=(x−2)2−4

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