# QUADRATIC FUNCTIONS : EXERCISE (5.3) SOLUTIONS

1.         Find the discriminant of each of the following quadratic functions. Also find the number of $x$ -intercepts of each of the functions.

$\begin{array}{l} \text{(a)}\ \ y=3 x^{2}-4 x+3\\\\ \text{(b)}\ \ y=2 x^{2}-4 x-3\\\\ \text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4\\\\ \text{(d)}\ \ y=-x^{2}+6 x-9\\\\ \text{(e)}\ \ y=-3 x^{2}-12 x-7\\\\ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 \end{array}$

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

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$\begin{array}{l} \ \ \ \ \ y=3{{x}^{2}}-4x+3\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=3,\ b=-4,\ c=3\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-4)}^{2}}-4(3)(3)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-20<0\\\\ \ \ \ \ \ \therefore \ \ \text{There is no }x-\text{intercept point.} \end{array}$

$\text{(b)}\ \ y=2 x^{2}-4 x-3$

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$\begin{array}{l} \ \ \ \ \ y=2{{x}^{2}}-4x-3\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=2,\ b=-4,\ c=-3\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-4)}^{2}}-4(2)(-3)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

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

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$\begin{array}{l}\ \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+x-4\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=\displaystyle \frac{1}{2},\ b=1,\ c=-4\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(1)}^{2}}-4\left( {\displaystyle \frac{1}{2}} \right)(-4)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

$\text{(d)}\ \ y=-x^{2}+6 x-9$

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$\begin{array}{l} \ \ \ \ \ y=-{{x}^{2}}+6x-9\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=-1,\ b=6,\ c=-9\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(6)}^{2}}-4(-1)(-9)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0\\\\ \ \ \ \ \ \therefore \ \ \text{There is one }x-\text{intercept point.}\ \end{array}$

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

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$\begin{array}{l} \ \ \ \ \ y=-3{{x}^{2}}-12x-7\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=-3,\ b=-12,\ c=-7\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-12)}^{2}}-4\left( {-3} \right)(-7)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =60>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

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

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$\begin{array}{l} \ \ \ \ \ y=-\displaystyle \frac{1}{2}{{x}^{2}}-3x-4\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=-\displaystyle \frac{1}{2},\ b=-3,\ c=-4\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-3)}^{2}}-4\left( {-\displaystyle frac{1}{2}} \right)(-4)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

2.         Find the intercept form of each of the quadratic functions. Also find the $y$ -intercept, axis of symmetry, vertex, and range of each of the functions.

$\begin{array}{l} \text{(a)}\ \ y=2 x^{2}-2 x-12\\\\ \text{(b)}\ \ y=3 x^{2}-6 x+3\\\\ \text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4\\\\ \text{(d)}\ \ y=2 x^{2}-5 x-3\\\\ \text{(e)}\ \ y=-6 x^{2}-7 x+5\\\\ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 \end{array}$

$\text{(a)}\ \ y=2 x^{2}-2 x-12$

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$\begin{array}{l} \ \ \ \ y=2{{x}^{2}}-2x-12\\\\ \ \ \ \ y=2({{x}^{2}}-x-6)\\\\ \ \ \ \ y=2(x+2)(x-3)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-12\\\\ \ \ \ \ \therefore \ y-\text{intercept point}=(0,\ -12)\\\\ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=2,\ p=-2,\ q=3\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{p+q}}{2}\Rightarrow x=\displaystyle \frac{1}{2}\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{1}{2},\ y=2{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)-12=-\displaystyle \frac{{25}}{2}\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {\displaystyle \frac{1}{2},-\displaystyle \frac{{25}}{2}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\ge -\displaystyle \frac{{25}}{2}} \right\} \end{array}$

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

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$\begin{array}{l} \ \ \ \ y=3{{x}^{2}}-6x+3\;\\\\ \ \ \ \ y=3({{x}^{2}}-2x+1)\\\\ \ \ \ \ y=3(x-1)(x-1)\\\\ \ \ \ \ \text{When}\ x=0,\ y=3\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ 3)\\\\ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=3,\ p=1,\ q=1\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{1+1}}{2}\Rightarrow x=1\\\\ \ \ \ \ \text{When}\ x=1,\ y=3(1-1)(1-1)=0\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {1,0} \right) \end{array}$

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

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$\begin{array}{l} \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+x-4\\\\ \ \ \ \ y=\displaystyle \frac{1}{2}({{x}^{2}}+2x-8)\\\\ \ \ \ \ y=\displaystyle \frac{1}{2}(x+4)(x-2)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-4\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ -4)\\\\ \text{ }\ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=\displaystyle \frac{1}{2},\ p=-4,\ q=2\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-4+2}}{2}\Rightarrow x=-1\\\\ \ \ \ \ \text{When}\ x=-1,\ y=\displaystyle \frac{1}{2}(-1+4)(-1-2)=-\displaystyle \frac{9}{2}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {-1,-\displaystyle \frac{9}{2}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\ge -\displaystyle \frac{9}{2}} \right\} \end{array}$

$\text{(d)}\ \ y=2 x^{2}-5 x-3$

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$\begin{array}{l} \ \ \ \ y=2{{x}^{2}}-5x-3\\\\ \ \ \ \ y=2\left( {{{x}^{2}}-\displaystyle \frac{5}{2}x-\displaystyle \frac{3}{2}} \right)\\\\ \ \ \ \ y=2\left( {x+\displaystyle \frac{1}{2}} \right)\left( {x-3} \right)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-3\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ -3)\\\\ \text{ }\ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=2,\ p=-\displaystyle \frac{1}{2},\ q=3\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-\displaystyle \frac{1}{2}+3}}{2}\Rightarrow x=\displaystyle \frac{5}{4}\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{5}{4},\ y=2\left( {\displaystyle \frac{5}{4}+\displaystyle \frac{1}{2}} \right)\left( {\displaystyle \frac{5}{4}-3} \right)=-\displaystyle \frac{{49}}{8}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {\displaystyle \frac{5}{4},-\displaystyle \frac{{49}}{8}} \right)\\\\\ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\ge -\displaystyle \frac{{49}}{8}} \right\} \end{array}$

$\text{(e)}\ \ y=-6 x^{2}-7 x+5$

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$\begin{array}{l}\ \ \ \ y=-6{{x}^{2}}-7x+5\\\\ \ \ \ \ y=-\left( {6{{x}^{2}}-7x-5} \right)\\\\ \ \ \ \ y=-(3x+5)(2x-1)\\\\\ \ \ \ y=-3\left( {x+\displaystyle \frac{5}{3}} \right)2\left( {x-\displaystyle \frac{1}{2}} \right)\\\\ \ \ \ \ y=-6\left( {x+\displaystyle \frac{5}{3}} \right)\left( {x-\displaystyle \frac{1}{2}} \right)\\\\ \ \ \ \ \text{When}\ x=0,\ y=5\\\\\ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ 5)\\\\ \ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=-6,\ p=-\displaystyle \frac{5}{2},\ q=\displaystyle \frac{1}{2}\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-\displaystyle \frac{5}{3}+\displaystyle \frac{1}{2}}}{2}\Rightarrow x=-\displaystyle \frac{7}{{12}}\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{5}{4},\ y=-6\left( {-\displaystyle \frac{7}{{12}}+\displaystyle \frac{5}{3}} \right)\left( {-\displaystyle \frac{7}{{12}}-\displaystyle \frac{1}{2}} \right)=\displaystyle \frac{{169}}{{24}}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {-\displaystyle \frac{7}{{12}},\displaystyle \frac{{169}}{{24}}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\le \displaystyle \frac{{169}}{{24}}} \right\} \end{array}$

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

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$\begin{array}{l} \ \ \ \ y=-\displaystyle \frac{1}{2}{{x}^{2}}-3x-4\\\\ \ \ \ \ y=-\displaystyle \frac{1}{2}\left( {{{x}^{2}}+6x+8} \right)\\\\ \ \ \ \ y=-\displaystyle \frac{1}{2}(x+4)(x+2)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-4\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ -4)\\\\ \ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=-\displaystyle \frac{1}{2},\ p=-4,\ q=-2\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-4-2}}{2}\Rightarrow x=-3\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{5}{4},\ y=-\displaystyle \frac{1}{2}(-3+4)(-3+2)=\displaystyle \frac{1}{2}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {-3,\displaystyle \frac{1}{2}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\le \displaystyle \frac{1}{2}} \right\} \end{array}$