Chapter 7.6, Problem 24WE

To determine

The maximum or minimum value of the function $h\left(x\right)=\left(2x-5\right)\left(2x+3\right)$ and then give the coordinates of the vertex of its graph.

Answer to Problem 24WE

The function $h\left(x\right)=\left(2x-5\right)\left(2x+3\right)$ has minimum value of $-16$ and the vertex of graph of function is $\left(\frac{1}{2},-16\right)$ .

Explanation of Solution

Given information: The given function is $h\left(x\right)=\left(2x-5\right)\left(2x+3\right)$ which can also be written as

  $h\left(x\right)=4x^2-4x-15\mathrm{....}\left(1\right)$

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, $a=4$ which is greater than zero. Therefore the given function will give the minimum value.

Now the value for $\left(x\right)$ associated with minimum value can be evaluated from the formula

  $x=-\frac{b}{2a}$ .

  $\begin{array}{c}x=-\frac{\left(-4\right)}{2\left(4\right)}\mathrm{....}\left(a=4,b=-4\right)\\ =\frac{1}{2}\end{array}$

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

  $\begin{array}{c}{\left[h\left(\frac{1}{2}\right)\right]}_{\min }=4{\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)-\mathrm{15....}\left(\text{From }\left(1\right)\right)\\ =-16\end{array}$

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

  $\begin{array}{c}y=4x^2-4x-\mathrm{15....}\left(\text{Replace }h\left(x\right)=y\right)\\ \text{or, }y+15=4\left(x^2-x\right)\\ y+15+\left(1\right)=4\left(x^2-2\left(x\right)\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2\right)\mathrm{....}\left(\text{Complete square}\right)\\ y+16=4{\left(x-\frac{1}{2}\right)}^2\mathrm{....}\left(3\right)\end{array}$

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 $\left(\frac{1}{2},-16\right)$ .