Question

Will Give Medal

f(x)=4x^2-8x+2 Write this function in vertex form (Please show me the steps)

Answer 1

The vertex form is like this: \[f(x)=a(x-h)^{2}+k\] So get there we need to complete the square on the quadratic equation.

Step 1:Set f(x)=0

Step 2: We need to clear the constant in front of the x^2 term. So divide the quadratic by 4\[(4)(x ^{2}-2x+2/4)=0\]
Step 3: Take 1/2 of the second term and square it. \[(-2*1/2)^{2}=1\]

Step 4: Insert the value in step 3 and it's opposite (-1) into the quadratic equation.\[(4)(x ^{2}-2x+1-1+1/2)=0\]

Step 5: Write \[x ^{2}-2x+1 =(x-1)^{2}\]

Step 6: Substitute and simplify.\[(4)((x-1)^{2}-1/2)=0\]\[f(x)=4(x-1)^{2}-2\]