## anonymous one year ago 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

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1. SolomonZelman

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)

2. SolomonZelman

So, you can tell that: $$\large\color{black}{ \displaystyle r-1=2t }$$ $$\large\color{black}{ \displaystyle r^2-r+1-4tr-4t=r-4t^2 }$$

3. SolomonZelman

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

4. anonymous

Thanks a lot @SolomonZelman very helpful