Quadratic Functions : Exercise (5.2) Solutions



1.         Find the vertex form of each of the following quadratic functions. Find also $y$ -intercept, axis of symmetry, vertex, and range of each of the functions.

           $\begin{array}{l} \text{(a)}\ \ y=2 x^{2}+4 x+3\\\\ \text{(b)}\ \ y=3 x^{2}-6 x+2\\\\ \text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4\\\\ \text{(d)}\ \ y=-2 x^{2}+2 x+3\\\\ \text{(e)}\ \ y=-3 x^{2}-12 x-7\\\\ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 \end{array}$


           $ \text{(a)}\;\;y=2{{x}^{2}}+4x+3$

Show/Hide Solution



$ \begin{array}{l} \ \ \ \ \ y=2{{x}^{2}}+4x+3\\\\ \ \ \ \ \ y=2\left( {{{x}^{2}}+2x+1} \right)+1\\\\ \ \ \ \ \ y=2{{\left( {x+1} \right)}^{2}}+1\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=3\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ 3).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=2,\ h=-1,\ k\ =1\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry:} x=h\Rightarrow x=-1\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(-1,1)\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \ge \ k\}=\{y\ |\ y\ \ge \ 1\} \end{array}$

           $\text{(b)}\;\;y=3{{x}^{2}}-6x+2$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ y=3{{x}^{2}}-6x+3-1\\\\ \ \ \ \ \ y=3\left( {{{x}^{2}}-2x+1} \right)-1\\\\ \ \ \ \ \ y=3{{\left( {x-1} \right)}^{2}}-1\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=2\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ 2).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=3,\ h=1,\ k\ =-1\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry: }x=h\Rightarrow x=1\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(1,-1)\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \ge \ k\}=\{y\ |\ y\ \ge \ -1\}\end{array}$

           ${\text{(c)}\;\;y=\displaystyle\frac{1}{2}{{x}^{2}}+x-4}$

Show/Hide Solution


$\begin{array}{l} \ \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+x-4\\\\ \ \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+\displaystyle \frac{2}{2}x+\displaystyle \frac{1}{2}-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ y=\displaystyle \frac{1}{2}\left( {{{x}^{2}}+2x+1} \right)-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ y=\displaystyle \frac{1}{2}{{\left( {x+1} \right)}^{2}}-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=-4\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ -4).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=\displaystyle \frac{1}{2},\ h=-1,\ k\ =-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry: }x=h\Rightarrow x=-1\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(-1,-\displaystyle \frac{9}{2})\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \ge \ k\}=\{y\ |\ y\ \ge \ -\displaystyle \frac{9}{2}\} \end{array}$

           $ \text{(d)}\;\ y=-2{{x}^{2}}+2x+3$

Show/Hide Solution


$\begin{array}{l} \ \ \ \ \ y=-2{{x}^{2}}+2x+3\\\\ \ \ \ \ \ y=-2\left( {{{x}^{2}}-x} \right)+3\\\\ \ \ \ \ \ y=-2\left( {{{x}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)x+\displaystyle \frac{1}{4}} \right)+3+\displaystyle \frac{1}{2}\\\\ \ \ \ \ \ y=-2{{\left( {x-\displaystyle \frac{1}{2}} \right)}^{2}}+\displaystyle \frac{7}{2}\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=3\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ 3).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=-2,\ h=\displaystyle \frac{1}{2},\ k\ =\displaystyle \frac{7}{2}\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry: }x=h\Rightarrow x=\displaystyle \frac{1}{2}\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(\displaystyle \frac{1}{2},\displaystyle \frac{7}{2})\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \le \ k\}=\{y\ |\ y\ \le \ \displaystyle \frac{7}{2}\} \end{array}$

           ${\text{(e)}\;\;y=-3{{x}^{2}}-12x-7}$

Show/Hide Solution


$\begin{array}{*{20}{l}} {\;\;\;\;\;y=-3{{x}^{2}}-12x-7} \\\\ {\;\;\;\;\;y=-3\left( {{{x}^{2}}+4x} \right)-7}\\\\ {\;\;\;\;\;y=-3\left( {{{x}^{2}}+4x+4} \right)-7+12}\\\\ {\;\;\;\;\;y=-3{{{\left( {x+2} \right)}}^{2}}+5} \\\\ {\;\;\;\;\;\text{When}\;x=0,\;y=-7}\\\\ {\;\;\;\;\;\therefore \;y-\text{intercept}\;(0,\;-7).}\\\\ {\;\;\;\;\;\text{Comparing with }a{{{\left( {x-h} \right)}}^{2}}+k,}\\\\ {\;\;\;\;\;a=-3,\;h=-2,\;k\;=5} \\ {} \\ {\;\;\;\;\;\therefore \;\text{axis of symmetry: }x=h\Rightarrow x=-2} \\\\ {\;\;\;\;\;\;\;\text{vertex}=(h,k)=(-2,5)}\\\\ {\;\;\;\;\;\;\;\text{range}=\{y\;|\;y\;\le \;k\}=\{y\;|\;y\;\le \;5\}} \end{array}$

           ${\text{(f)}\;\;y=-\displaystyle\frac{1}{2}{{x}^{2}}-3x-4}$

Show/Hide Solution




$\begin{array}{*{20}{l}} {\;\;\;\;\;y=-\displaystyle \frac{1}{2}{{x}^{2}}-3x-4} \\ {} \\ {\;\;\;\;\;y=-\displaystyle \frac{1}{2}\left( {{{x}^{2}}+6x} \right)-4} \\ {} \\ {\;\;\;\;\;y=-\displaystyle \frac{1}{2}\left( {{{x}^{2}}+2(3)x+9} \right)-4+\displaystyle \frac{9}{2}} \\ {} \\ {\;\;\;\;\;y=-\displaystyle \frac{1}{2}{{{\left( {x+3} \right)}}^{2}}+\displaystyle \frac{1}{2}} \\ {} \\ {\;\;\;\;\;\text{When}\;x=0,\;y=-4} \\ {} \\ {\;\;\;\;\;\therefore \;y-\text{intercept}\;(0,\;-4).} \\ {} \\ {\;\;\;\;\;\text{Comparing with }a{{{\left( {x-h} \right)}}^{2}}+k,} \\ {} \\ {\;\;\;\;\;a=-\displaystyle \frac{1}{2},\;h=-3,\;k\;=\displaystyle \frac{1}{2}} \\ {} \\ {\;\;\;\;\;\therefore \;\text{axis of symmetry: }x=h\Rightarrow x=-3} \\ {} \\ {\;\;\;\;\;\;\;\text{vertex}=(h,k)=(-3,\displaystyle \frac{1}{2})} \\ {} \\ {\;\;\;\;\;\;\;\text{range}=\{y\;|\;y\;\le \;k\}=\{y\;|\;y\;\le \;\displaystyle \frac{1}{2}\}} \end{array}$


2.         Find two positive numbers whose sum is $20$ and whose product is as large as possible.

Show/Hide Solution

Let the two positive numbers be $x$ and $y$.

By the problem,

$x + y = 20$

$\therefore\ y=20-x$

Let the product of the numbers be $p$.

Hence,

$\begin{aligned} p &=xy\\\\ &=x(20-x)\\\\ &=20x-x^2\\\\ &=-(x^2-20x)\\\\ &=-(x^2-20x + 100 )+100\\\\ &=100-(x-10)^2\\\\ \therefore\ p&\le100 \end{aligned}$

Thus, maximum value of $p$ is $100$ when $(x-10)^2 = 0$.

$\therefore\ x = 10$.

When $x=10, y = 20 - x = 20 - 10 = 10$.

Therefore, the twopositive numbers are $x=10$ and $y=10$.


3.         What is the largest area possible for a rectangle whose perimeter is $16$ cm?

Show/Hide Solution

Let the lenght and breadth of the rectangle be $x$ and $y$ respectively.

By the problem,

$2x + 2y = 16$

$\therefore\ y=8-x$

Let the area of the rectangle be $A$.

Hence,

$\begin{aligned} A &=xy\\\\ &=x(8-x)\\\\ &=8x-x^2\\\\ &=-(x^2-8x)\\\\ &=-(x^2-2\cdot 4\cdot x + 16 )+16\\\\ &=16-(x-4)^2\\\\ \therefore\ A&\le 16 \end{aligned}$

Thus, the maximum area $(A)$ of the rectangle is $16$ cm².