Answer:
[tex] f(t) = 4(t-1)^2 +4[/tex]
So then we see that the required values are:
[tex] a = 4, h =1, k =4[/tex]
And the vertex for this case would be:
[tex] V_x = -\frac{-8}{2*4} = 1[/tex]
[tex] V_y = 4[/tex]
[tex] V(1,4)[/tex]
Step-by-step explanation:
For this case we have the following function given:
[tex] f(t) = 4t^2 -8t +8[/tex]
And we want to wrote this equation in the form a(x − h)2 + k
We can divide both sides of the equation and we got:
[tex] \frac{f(t)}{4} = t^2 -2t +2[/tex]
Now we can comple the square in the rigth part with this:
[tex]\frac{f(t)}{4} = t^2 -2t +1 +(2-1)[/tex]
[tex]\frac{f(t)}{4} = (t^2 -2t +1) +(2-1)= (t-1)^2 +1[/tex]
And now we can multiply both sides by 4 and we got:
[tex] f(t) = 4(t-1)^2 +4[/tex]
So then we see that the required values are:
[tex] a = 4, h =1, k =4[/tex]
And the vertex for this case would be:
[tex] V_x = -\frac{-8}{2*4} = 1[/tex]
[tex] V_y = 4[/tex]
[tex] V(1,4)[/tex]