quadratic function in the form f(x)=a(x-h)^2+k. Then transformations graph f(x)=x^2

Question

2. f(x)=1/4x^2-2x+3

 

3. f(x)=-1/2x^2+4x-2

Answer 1

Expanding the standard form: f(x)=ax^2-2ahx+ah^2+k and comparing terms we get: a=1/4; -2ah=-2, so h=1/a=4; and ah^2+k=3, making k=3-ah^2=3-16/4=-1. So (2) f(x)=(1/4)(x-4)^2-1 in standard form or (y+1)=(x-4)^2/4, where y=f(x).

Similarly a=-1/2, -2ah=4, so h=4 and ah^2+k=-2, making k=-2+16/2=6. So (3) f(x)=-1/2(x-4)^2+6 or (y-6)=-(x-4)^2, where y=f(x).

The origin of parabola (2) is (h,k)=(4,-1) and (3) is (4,6). Transformation, or shift, of the origin to these points gives (2) f(x)=x^2/4 and (3) f(x)=-x^2/2.