Question

Use the method of symmetry to find the extreme value of each quadratic function and the value of $x$ for which it occurs. $g\left(x\right)=(5-x)(2x+3)$ pls give it in this form:

The $x$-intercepts of the graph of the function are (_,_),(_,_). The midpoint of the $x$-intercepts is ___. The extreme value is (a maximum or a minimum).

$g\left(\right)$=_____

Answer 1

Step-1: Find the $x$-intercepts:

Given a function $g\left(x\right)=(5-x)(2x+3)$.

The x-intercepts of a function is value where the corresponding y-coordinate is $0.$

Therefore the function can be given as follows:

$(5-x)(2x+3)=0$

Therefore on solving and simplifying the equation.

$$\begin{gathered} 5-x=0 \\ \Rightarrow x=5 \end{gathered}$$

And similarly for the other term,

$$\begin{gathered} 2x+3=0 \\ \Rightarrow 2x=-3 \\ \Rightarrow x=-\frac{3}{2} \end{gathered}$$

The coordinates of the $x$-intercepts are $(5,0)\mathrm{and}\left(-\frac{3}{2},0\right)$

Step-2: Find the midpoint of the $x$-intercept:

Now the coordinates of the $x$-intercepts are $(5,0)\mathrm{and}\left(-\frac{3}{2},0\right)$.

The midpoint of two points $(a,b)\mathrm{and}\mathit{(}c\mathit{,}d\mathit{)}$ can be given as follows

$\mathrm{Midpoint}=\left(\frac{\mathrm{a}+\mathrm{c}}{2},\frac{\mathrm{b}+\mathrm{d}}{2}\right)$

Substituting the values in the midpoint formula and on simplify to get the midpoint.

$$\begin{gathered} \mathrm{Midpoint}=\left(\frac{5-{\displaystyle \frac{3}{2}}}{2},0\right) \\ \mathrm{Midpoint}=\left(\frac{7}{4},0\right) \end{gathered}$$

Hence the midpoint of the $x$-intercepts is $\frac{7}{4}$

Step-3: Find the extreme value:

Expand the given equation.

$$\begin{gathered} g\left(x\right)=(5-x)(2x+3) \\ g\left(x\right)=10x+15-2x^2-3x \\ g\left(x\right)=-2x^2+7x+15 \end{gathered}$$

Comparing the equation with standard form of quadratic equation $a{x}^2+bx+c=0$

Now the extreme value occurs at $x=-\frac{b}{2a}$.

Computing this value

$$\begin{gathered} -\frac{b}{2a}=-\frac{7}{2(-2)} \\ \Rightarrow -\frac{b}{2a}=\frac{7}{4} \end{gathered}$$

Now substitute this value in the original function.

$$\begin{gathered} g\left(x\right)=(5-x)(2x+3) \\ g\left(\frac{7}{4}\right)=\left(5-\frac{7}{4}\right)\left(\frac{7}{2}+3\right) \\ g\left(\frac{7}{4}\right)=\left(\frac{13}{4}\right)\left(\frac{13}{2}\right) \\ g\left(\frac{7}{4}\right)=\left(\frac{169}{8}\right) \end{gathered}$$

Hence, The $x$-intercepts are $\mathbf{5}\mathbf{,}\mathbf{}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{.}$ The midpoint of the $x$-intercepts are $\frac{\mathbf{7}}{\mathbf{4}}.$ The extreme value of the function $\mathit{g}\left(\frac{7}{4}\right)\mathbf{=}\left(\frac{169}{8}\right)\mathbf{.}$