Question

the quadratic function f(x)=x^2 -4tx+r has a minimum value 4t-4t^2 , where t and r are constants. the graph of function symmetrical about x= r-1.
by using complete the square find value of r and t

Answer 1

If the graph of the function is summetric about x=r-1, then the vertex is located at that line. (And vertex is a minimum if the parabola opens up.)
And
f(r-1)=(r-1)²-4t(r-1)+r
f(r-1)=(r²-2r+1)-4tr-4t+r
f(r-1)=r²-r+1-4tr-4t

So vertex is ( r-1, r²-r+1-4tr-4t)


\(\large\color{black}{ \displaystyle f(x)=x^2 -4tx+r }\)

\(\large\color{black}{ \displaystyle f(x)=x^2 -4tx+r+\left(\frac{4t}{2}\right)^2-\left(\frac{4t}{2}\right)^2 }\)

\(\large\color{black}{ \displaystyle f(x)=x^2 -4tx+\left(\frac{4t}{2}\right)^2+r-\left(\frac{4t}{2}\right)^2 }\)

\(\large\color{black}{ \displaystyle f(x)=x^2 -4tx+\left(2t \right)^2+r-\left(2t\right)^2 }\)

\(\large\color{black}{ \displaystyle f(x)=\left(x -2t\right)^2+r-4t^2 }\)

And according to completing the square, the vertex is:
(2t, r-4t²)

Answer 2

So, you can tell that:

\(\large\color{black}{ \displaystyle r-1=2t }\)
\(\large\color{black}{ \displaystyle r^2-r+1-4tr-4t=r-4t^2 }\)

Answer 3

that was based on the vertex....
Now, just solve the system.

Answer 4

Thanks a lot @SolomonZelman
very helpful