Wednesday, July 14, 2021

Exercise (5.1)- Translation of $y=x^2$

  1. Compare the graphs of the following functions to the graph of $y=x^{2}$.

    (a) $y=x^{2}+2 x+3$


    $\begin{array}{l}y={{x}^{2}}+2x+3\\\\\ \ \ \ \ ={{x}^{2}}+2x+1+2\\\\\ \ \ \ \ ={{(x+1)}^{2}}+2\\\\\ \ \ \ \ ={{\left( {x-(-1)} \right)}^{2}}+2\end{array}$


    Therefore, the graph of $y=x^{2}+2 x+3$ is the translation of negative 1 unit horizontally and positive 2 units vertically of the graph $y=x^{2}$.


    (b) $y=x^{2}-4 x+2$


    $\begin{array}{l}\ \ y\ ={{x}^{2}}-4x+2\\\\\ \ \ \ \ ={{x}^{2}}-4x+4-2\\\\\ \ \ \ \ ={{(x-2)}^{2}}-2\end{array}$


    Therefore, the graph of $y=x^{2}-4 x+2$ is the translation of positive 2 units horizontally and negative 2 units vertically of the graph $y=x^{2}$.


    (c) $y=x^{2}+4 x+4$


    $\begin{array}{l}\ \ y={{x}^{2}}+4x+4\\\\\ \ \ \ \ ={{(x+2)}^{2}}\end{array}$


    Therefore, the graph of $y=x^{2}+4 x+4$ is the translation of negative 2 units horizontally of the graph $y=x^{2}$ without vertical shift.

  2. Compare the graphs of the following functions to the graph of $y=-x^{2}$.

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


    $\begin{array}{l}\ \ y=-{{x}^{2}}+2x+3\\\\\ \ \ \ \ =-({{x}^{2}}-2x)+3\\\\\ \ \ \ \ =-({{x}^{2}}-2x+1)+3+1\\\\\ \ \ \ \ =-{{(x-1)}^{2}}+4\end{array}$


    Therefore, the graph of $y=-x^{2}+2 x+3$ is the translation of positive 1 unit horizontally and positive 4 units vertically of the graph $y=-x^{2}$.


    (b) $y=-x^{2}-4 x-7$


    $\begin{array}{l}\ \ y=-{{x}^{2}}-4x-7\\\\\ \ \ \ \ =-({{x}^{2}}+4x)-7\\\\\ \ \ \ \ =-({{x}^{2}}+4x+4)-7+4\\\\\ \ \ \ \ =-{{(x+2)}^{2}}-3\end{array}$


    Therefore, the graph of $y=-x^{2}-4 x-7$ is the translation of negative 2 units horizontally and negative 3 units vertically of the graph $y=-x^{2}$.


    (c) $y=-x^{2}+4 x-4$


    $\begin{array}{l}\ \ y=-{{x}^{2}}+4x-4\\\\\ \ \ \ \ =-({{x}^{2}}-4x+4)\\\\\ \ \ \ \ =-{{(x-2)}^{2}}\end{array}$


    Therefore, the graph of $y=-x^{2}+4 x-4$ is the translation of positive 2 units horizontally of the graph $y=-x^{2}$ without vertical shift.

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