C cpags68 New member Joined Mar 25, 2021 Messages 1 Mar 25, 2021 #1 Given f(x)=ax^2+bx+c, with a not equal to 0; the equation f(x)=5 has exactly one solution, x=2; and f(5)=8. Find the vertex of the graph y=f(x).
Given f(x)=ax^2+bx+c, with a not equal to 0; the equation f(x)=5 has exactly one solution, x=2; and f(5)=8. Find the vertex of the graph y=f(x).
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 Mar 25, 2021 #2 the vertex form of a quadratic function is [MATH]f(x) = a(x-h)^2 + k[/MATH] where [MATH](h,k)[/MATH] is the vertex. Using that form, what would be the function value of [MATH]f(h)[/MATH]? Would any other value of x yield that same value?
the vertex form of a quadratic function is [MATH]f(x) = a(x-h)^2 + k[/MATH] where [MATH](h,k)[/MATH] is the vertex. Using that form, what would be the function value of [MATH]f(h)[/MATH]? Would any other value of x yield that same value?
