The quadratic $ax^2 + bx + c$ can be expressed in the form $2(x - 4)^2 + 8$. When the quadratic $3ax^2 + 3bx + 3c$ is expressed in the form $n(x - h)^2 + k$, what is $h$?
Question:
The quadratic \(ax^2+bx+c\) can be expressed in the form \(2(x-4)^2+8\). When the quadratic \(3ax^2+3bx+3c\) is expressed in the form \(n(x-h)^2+k\), what is h?
Well if the quadratic \(ax^2+bx+c\) is the same as \(2(x-4)^2+8\) then we can say they are congruent, hence identical.
That is:
\(ax^2+bx+c =2(x-4)^2+8 \\ ax^2+bx+c=2(x^2-8x+16)+8 \\ ax^2+bx+c=2x^2-16x+40 \)
Hence, we see that:
\(a=2 \\ b=-16 \\ c=40\)
Substitute these in: \(3ax^2+3bx+3c\)
\(3(2)x^2+3(-16)x+3(40)=6x^2-48x+120\)
Now we want to express this in the form \(n(x-h)^2+k\)
We use the same technique:
\(6x^2-48x+120=n(x^2-2xh+h^2)+k\)
So that:
\(6x^2-48x+120=nx^2-2xnh+nh^2+k\)
We see that:
\(6=n \\ 48 = 2nh \implies 48=2(6)h \implies h=4\)
Which is what we wanted. I hope this helps!
