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This is one of our many quizzes on Completing the square to vertex form. Click on this to access more such competitive quizzes on Conic Sections.

Completing the square to vertex form Quiz 2

mathematics

 Find the vertex of the equation $y=3{{x}^{2}}+5x-4$.

 Apply $h=-\dfrac{b}{2a}$, $k=f\left( h \right)$ into $y=a{{\left( x-h \right)}^{2}}+k$ to get the vertex from equation.

  $\begin{align}  & \text{Given }y=3{{x}^{2}}+5x-4 \\  & \therefore h=-\dfrac{5}{6} \\  & \Rightarrow k=3\times \dfrac{25}{36}+5\times \left( -\dfrac{5}{6} \right)-4 \\  & \Rightarrow \dfrac{25}{12}-\dfrac{25}{6}-\dfrac{4}{1} \\  & \Rightarrow \dfrac{25-50-48}{12} \\  & \Rightarrow -\dfrac{73}{12} \\  & \text{Now} \\  & \Rightarrow y=3{{\left( x+\dfrac{5}{6} \right)}^{2}}+\left( -\dfrac{73}{12} \right) \\  & \Rightarrow y+\dfrac{73}{12}=3\dfrac{{{\left( 6x+5 \right)}^{2}}}{36} \\  & \Rightarrow 12y+73={{\left( 6x+5 \right)}^{2}} \\ \end{align}$

 If the value of $k=120$ then what is the value of $'c'$ if quadratic equation $y=2{{x}^{2}}-5x+c$.

 Firstly calculate the value of $h$ and then further solve.

  $\begin{align}  & \text{Given }y=2{{x}^{2}}-5x+c \\  & \Rightarrow h=-\dfrac{\left( -5 \right)}{4} \\  & \Rightarrow h=\dfrac{5}{4} \\  & \Rightarrow k=2{{\left( \dfrac{5}{4} \right)}^{2}}-5\times \left( \dfrac{5}{4} \right)+c \\  & \Rightarrow 120=2\times \dfrac{25}{16}-\dfrac{25}{4}+c \\  & \Rightarrow 120=\dfrac{25-50}{8}+c \\  & \Rightarrow c=\dfrac{120}{1}+\dfrac{25}{8} \\  & \Rightarrow c=\dfrac{960+25}{8} \\  & \Rightarrow c=123.125 \\ \end{align}$

 Find the value $'h'$ and $'k'$ if quadrant equation $y=\sqrt{3}{{x}^{2}}+\log 2x+5$.

 Apply $\log 2=0.30$ for the getting the value of $h$ and $k$.

$\begin{align}  & \text{Given }y=\sqrt{3}{{x}^{2}}+\log 2x+5 \\  & \Rightarrow h=-\dfrac{\log 2}{2\sqrt{3}} \\  & \Rightarrow h=-\dfrac{0.30}{2\sqrt{3}} \\  & \Rightarrow k=\sqrt{3}\times {{\left( \dfrac{0.30}{2\sqrt{3}} \right)}^{2}}+\log 2\times \dfrac{0.30}{2\sqrt{3}}+5 \\  & \Rightarrow k=\sqrt{3}\times \dfrac{0.09}{12}+0.30\times \left( -\dfrac{0.30}{2\sqrt{3}} \right)+5 \\  & \Rightarrow k=\dfrac{0.09\sqrt{3}}{12}-\dfrac{0.09}{2\sqrt{3}}+\dfrac{5}{1} \\  & \Rightarrow k=\dfrac{0.09\sqrt{3}-0.09\times 2\sqrt{3}+5\times 12}{12} \\  & \Rightarrow k=\dfrac{0.09\sqrt{3}-0.18\sqrt{3}+60}{12} \\  & \Rightarrow k=\dfrac{60-0.09\sqrt{3}}{12} \\ \end{align}$

 Find quadrant equation if vertex $5y+52=25{{\left( x+\dfrac{3}{5} \right)}^{2}}$.

 Firstly calculate the value of $h$ and $k$ then further solve for the quadratic equation.

 $\begin{align}  & \text{Given }5y+52=25{{\left( x+\dfrac{3}{5} \right)}^{2}} \\  & \Rightarrow y-\dfrac{52}{5}=5{{\left( x+\dfrac{3}{5} \right)}^{2}} \\  & \Rightarrow y=5{{\left( x+\dfrac{3}{5} \right)}^{2}}-\dfrac{52}{5} \\  & \Rightarrow y=a{{\left( x-h \right)}^{2}}+k \\  & \therefore a=5 \\  & \Rightarrow h=\dfrac{3}{5} \\  & \Rightarrow -\dfrac{b}{2a}=\dfrac{3}{5} \\  & \Rightarrow -\dfrac{b}{2\times 5}=\dfrac{3}{5} \\  & \Rightarrow -b=\dfrac{3}{5}\times 2\times 5 \\  & \Rightarrow b=-6 \\  & \Rightarrow k=5{{\left( \dfrac{3}{5} \right)}^{2}}-6\left( \dfrac{3}{5} \right)+c \\  & \Rightarrow -\dfrac{52}{5}=\dfrac{9}{5}-\dfrac{18}{5}+c \\  & \Rightarrow c=\dfrac{-52+9}{5} \\  & \Rightarrow c=-\dfrac{43}{5} \\  & \therefore y=5{{x}^{2}}-6x-\dfrac{43}{5} \\ \end{align}$

 Find the vertex equation $y=\sqrt{3}{{x}^{2}}+2\sqrt{5}x+5$.

 Apply $h=-\dfrac{b}{2a}$, $k=f\left( h \right)$ then apply $y=a{{\left( x-h \right)}^{2}}+k$.

 $\begin{align}  & \text{Given }y=\sqrt{3}{{x}^{2}}+2\sqrt{5}x+5 \\  & \Rightarrow h=-\dfrac{2\sqrt{5}}{2\sqrt{3}} \\  & \Rightarrow h=\dfrac{\sqrt{5}}{\sqrt{3}} \\  & k=f\left( h \right) \\  & \therefore k=\sqrt{3}\times \dfrac{5}{3}+2\sqrt{5}\times \dfrac{\sqrt{5}}{\sqrt{3}}+5 \\  & \Rightarrow k=\dfrac{5}{\sqrt{3}}+\dfrac{10}{\sqrt{3}}+\dfrac{5}{1} \\  & \Rightarrow k=\dfrac{5+10+5\sqrt{3}}{\sqrt{3}} \\  & \Rightarrow k=\dfrac{15+5\sqrt{3}}{\sqrt{3}} \\  & \therefore y=\sqrt{3}{{\left( x-\dfrac{\sqrt{5}}{\sqrt{3}} \right)}^{2}}+\dfrac{15+5\sqrt{3}}{\sqrt{3}} \\  & \Rightarrow y-\left( \dfrac{15+5\sqrt{3}}{\sqrt{3}} \right)=\sqrt{3}{{\left( x-\dfrac{\sqrt{5}}{\sqrt{3}} \right)}^{2}} \\  & \Rightarrow \dfrac{\sqrt{3}y-\left( 15+5\sqrt{3} \right)}{\sqrt{3}}=\sqrt{3}{{\left( x-\dfrac{\sqrt{5}}{\sqrt{3}} \right)}^{2}} \\  & \Rightarrow \sqrt{3}y-\left( 15+5\sqrt{3} \right)=3{{\left( \dfrac{\sqrt{3}x-\sqrt{5}}{3} \right)}^{2}} \\  & \Rightarrow \sqrt{3}y-\left( 15+5\sqrt{3} \right)={{\left( \sqrt{3}x-\sqrt{5} \right)}^{2}} \\ \end{align}$