Chapter 1, Problem 12TP

Solution

To find: The maximum value of the function $y=-3{\left(x-2\right)}^2+6$ .

The maximum value of the function is $6$ .

Given:

The function is $y=-3{\left(x-2\right)}^2+6$ .

Concept used:

If $\left(m\right)$ is the maximum value of the function of the form $y=a{x}^2+bx+c$ where $a<0$ then maximum value of the function is written as:

  $m=c-\frac{b^2}{4a}\mathrm{......}\left(1\right)$

Calculation:

Re-write the function $y=-3{\left(x-2\right)}^2+6$ in the form of $y=a{x}^2+bx+c$ .

  $\begin{array}{c}y=-3{\left(x-2\right)}^2+6\mathrm{......}\left(\text{Given}\right)\\ y=-3\left(x^2+4-4x\right)+6\\ y=-3x^2+12x-6\end{array}$

Comparing the function $y=-3x^2+12x-6$ with $y=a{x}^2+bx+c$ where $a<0$ gives $a=-3,b=12\text{ and }c=-6$ .

Now, with the help of formula $\left(1\right)$ the maximum value $\left(m\right)$ of the function is evaluated as:

  $\begin{array}{c}m=c-\frac{b^2}{4a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =-6-\frac{{\left(12\right)}^2}{4\left(-3\right)}\mathrm{......}\left(\begin{array}{l}\because a=-3,b=12\\ \text{ and }c=-6\end{array}\right)\\ =6\end{array}$

Conclusion:

Hence, the maximum value of the function is $6$ .