Graphing Functions Using Vertical and Horizontal Shift Practice

Practice Problems

Answers

  1. The graph of function ‍h is the graph of function ‍f shifted ‍12 units to the right and down 4 units.

    To shift the graph of y=f(x) right ‍12 units, we can replace ‍x with x-12 in the equation.

    Two congruent absolute value functions labeled f of x and f of x minus twelve quantity on an x y coordinate plane. The x- and

    To shift the graph of y=f(x-12) down ‍4 units, we can replace ‍f(x-12) with f(x-12)-4 in the equation.

    Three congruent absolute value functions labeled f of x, f of x minus twelve quantity, and f of x minus twelve quantity minus

    Notice that this is the graph of function h, and so h(x)=f(x-12)-4.

    The answer:

    h(x)=f(x-12)-4


  2. The graph of ‍y=x^2 can be transformed to get the graph of ‍‍y=(x+4)^2+2.

    • Replacing ‍x with ‍‍x+4 shifts the graph of ‍‍y=x^2 to the left by ‍4 units.
    • Adding 2 to the function shifts the graph of ‍‍y=x^2 up by ‍2 units.


    Therefore, the graph of ‍y=(x+4)^2+2 is obtained by shifting the graph of ‍‍y=x^2 to the left by ‍‍4 units and up by‍ 2 units.

    The graph that corresponds with this transformation is graph B

    q2-answer - b


  3. The graph of ‍y=f(x) can be transformed to get the graph of ‍‍y=f(x-4)+2.

    • Replacing ‍x with ‍‍x-4 shifts the graph of ‍‍y=f(x) to the right by ‍4 units.

    • Adding 2 to the function shifts the graph of ‍‍y=f(x) up by ‍2 units.


    Therefore, the graph of ‍y=f(x-4)+2 is obtained by shifting the graph of ‍‍y=f(x) to the right by ‍‍4 units and up by‍ 2 units.

    The graph that corresponds with this transformation is graph D

    q3-anwer-d


  4. The graph of function ‍g is the graph of function ‍f shifted ‍left 2 units and up 5 units.

    To shift the graph of y=f(x) left 2 units, we can replace ‍x with x+2 in the equation.

    \begin{aligned}f({x})&=\sqrt{{x}+4}-2\\\\f({x+2})&=\sqrt{{x+2}+4}-2\\\\&=\sqrt{x+6}-2\end{aligned}

    To shift this graph up 5, we can add 5 to the function value:

    \begin{aligned}f({x+2})&=\sqrt{{x+6}}-2\\\\f(x+2)+{5}&=\sqrt{{x+6}}-2 + 5\\\\&=\sqrt{x+6}+3\end{aligned}

    The answer:

    g(x)=\sqrt{x+6}+3