Question 1005898
{{{y=kx^2-2kx+3}}}


{{{k<0}}} and {{{y=4}}} where x is the middle value between the zeros.


Finding zeros using formula for general solution of quadratic equation:
{{{x=(2k+- sqrt(4k^2-4k*3))/(2k)}}}
D for discriminant,
{{{x=(2k+- sqrt(D))/(2k)}}}



Middle x between the zeros is {{{(2k+2k)/(2*2k)}}}
{{{4k/4k}}}
{{{highlight(1)}}}
-
This means x=1 and y=4  is the maximum or "optimum" or VERTEX point for the function.


{{{4=k*1(1-2)+3}}} using the given original equation or function
{{{4=k(-1)+3}}}
{{{4=-k+3}}}
{{{-4=k-3}}}
{{{k=-4+3}}}
{{{highlight(k=-1)}}}