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