Chapter 7.6, Problem 23WE

To determine

The maximum or minimum value of the function $h\left(x\right)=\left(2-x\right)\left(5+x\right)$ and then give the coordinates of the vertex of its graph.

Answer to Problem 23WE

The function $h\left(x\right)=\left(2-x\right)\left(5+x\right)$ has maximum value of $\frac{49}{12}$ and the vertex of graph of function is $\left(-\frac{3}{2},\frac{49}{4}\right)$ .

Explanation of Solution

Given information: The given function is $h\left(x\right)=\left(2-x\right)\left(5+x\right)$ which can also be written as

  $h\left(x\right)=-x^2-3x+10\mathrm{....}\left(1\right)$

Formula used: The function of the form $f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ has the maximum value if $a<0$ and minimum value when $a>0$ . Mathematically,

  $\left(\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\text{ is maximum for }a<0\\ \text{ is minimum for }a>0\end{array}\right)\mathrm{....}\left(2\right)$

Calculation: Comparing the formula (2) with (1) gives, $a=-1$ which is less than zero. Therefore the given function will give the maximum value.

Now the value for $\left(x\right)$ associated with maximum value can be evaluated from the formula $x=-\frac{b}{2a}$ .

  $\begin{array}{c}x=-\frac{\left(-3\right)}{2\left(-1\right)}\mathrm{....}\left(a=-1,b=-3\right)\\ =-\frac{3}{2}\end{array}$

Hence the value of function will give the maximum value at $x=-\frac{3}{2}$ . Now evaluate the maximum value at $x=-\frac{3}{2}$ from (1).

  $\begin{array}{c}{\left[h\left(-\frac{3}{2}\right)\right]}_{\max }=-{\left(-\frac{3}{2}\right)}^2-3\left(-\frac{3}{2}\right)+\mathrm{10....}\left(\text{From }\left(1\right)\right)\\ =\frac{49}{12}\end{array}$

Again from (1) the given function can be written as,

  $\begin{array}{c}y=-x^2-3x+\mathrm{10....}\left(\text{Replace }h\left(x\right)=y\right)\\ \text{or, }y-10=-1\left(x^2+3x\right)\\ y-10+\left(-\frac{9}{4}\right)=-\left(x^2+2\left(x\right)\left(\frac{3}{2}\right)+{\left(\frac{3}{2}\right)}^2\right)\mathrm{....}\left(\text{Complete square}\right)\\ y-\frac{49}{4}=-{\left(x+\frac{3}{2}\right)}^2\mathrm{....}\left(3\right)\end{array}$

Since (3) represents a parabola of the form $y-k=a{\left(x-h\right)}^2$ . Therefore the graph for the function (1) has vertex $\left(-\frac{3}{2},\frac{49}{4}\right)$ .