91Ó°ÊÓ

Consider the function $f\left(x\right)=-4x^2+12x-5$.

The function's leading coefficient is negative so the parabola opens downward.

So the y coordinate of the vertex gives the maximum value of the function.

On comparing the given function $f\left(x\right)=-4x^2+12x-5$with the standard form $f\left(x\right)=a{x}^2+bx+c$we get

$$\begin{gathered} a=-4 \\ b=12 \\ c=-5 \end{gathered}$$

The axis of symmetry for the standard form is $x=-\frac{b}{2a}$.

So for the given function, the axis of symmetry is given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{12}{2·(-4)} \\ x=-\frac{3}{-2} \\ x=1.5 \end{gathered}$$

Thus the equation of the axis of symmetry is $x=1.5$.

Vertex is a point on the line of axis of symmetry.

Here axis of symmetry is $x=1.5$.

So the function value $f(1.5)$is given as

$\begin{array}{rcl}f(1.5)& =& -4·{(1.5)}^2+12·(1.5)-5\\ & =& -9+18-5\\ & =& 4\end{array}$

So the vertex of the parabola is $(-1.5,4)$.

And so the maximum value of the function is $4$.

The graph of the function with its vertex can be seen to verify the result.