Respuesta :
This is an equation of a parabola (quadratic) of the form:
[tex]ax^2+bx+c[/tex]According to the equation, the values of the constants are:
a = 6
b = -1
c = -4
The max value of a parabolic function occurs at:
[tex]x=-\frac{b}{2a}[/tex]We substitute and find:
[tex]\begin{gathered} x=-\frac{-1}{2(6)} \\ x=\frac{1}{12} \end{gathered}[/tex]Then, if u plug in that number into the function, you will get the max/min value:
[tex]\begin{gathered} f(\frac{1}{12})=6(\frac{1}{12})^2-\frac{1}{12}-4 \\ =6(\frac{1}{144})-\frac{1}{12}-4 \\ =\frac{1}{24}-\frac{1}{12}-4 \\ =\frac{1-1(2)-4(24)}{24} \\ =\frac{1-2-96}{24} \\ =-\frac{97}{24} \end{gathered}[/tex]So, the extrema value is at:
[tex](\frac{1}{12},-\frac{97}{24})[/tex]Also, this is a minimum.
If a > 0 , we have a minimum
If a < 0, we have a max.
Since a = 6 which is greater than 0, so the answer is minimum.
