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anonymous
 one year ago
the quadratic function f(x)=x^2 4tx+r has a minimum value 4t4t^2 , where t and r are constants. the graph of function symmetrical about x= r1.
by using complete the square find value of r and t
anonymous
 one year ago
the quadratic function f(x)=x^2 4tx+r has a minimum value 4t4t^2 , where t and r are constants. the graph of function symmetrical about x= r1. by using complete the square find value of r and t

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0If the graph of the function is summetric about x=r1, then the vertex is located at that line. (And vertex is a minimum if the parabola opens up.) And f(r1)=(r1)²4t(r1)+r f(r1)=(r²2r+1)4tr4t+r f(r1)=r²r+14tr4t So vertex is ( r1, r²r+14tr4t) \(\large\color{black}{ \displaystyle f(x)=x^2 4tx+r }\) \(\large\color{black}{ \displaystyle f(x)=x^2 4tx+r+\left(\frac{4t}{2}\right)^2\left(\frac{4t}{2}\right)^2 }\) \(\large\color{black}{ \displaystyle f(x)=x^2 4tx+\left(\frac{4t}{2}\right)^2+r\left(\frac{4t}{2}\right)^2 }\) \(\large\color{black}{ \displaystyle f(x)=x^2 4tx+\left(2t \right)^2+r\left(2t\right)^2 }\) \(\large\color{black}{ \displaystyle f(x)=\left(x 2t\right)^2+r4t^2 }\) And according to completing the square, the vertex is: (2t, r4t²)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0So, you can tell that: \(\large\color{black}{ \displaystyle r1=2t }\) \(\large\color{black}{ \displaystyle r^2r+14tr4t=r4t^2 }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0that was based on the vertex.... Now, just solve the system.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a lot @SolomonZelman very helpful
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