Question

match each verbal description to its equivelent function rule as applied to the given function below.

Answer 1

The parent function is f(x) = 7x + 5.

So let's find the equation for each transformation:

a) The function f reflected about the y-axis and translated 3 units left.

A reflection about the y-axis can be calculated just changing the sign of the variable x:

$f(x)=7x+5\to f^{\prime}(x)=7\cdot(-x)+5=-7x+5$

Now, in order to translate 3 units left, we just need to add 3 units to the x-value:

$\begin{gathered} f^{\prime}(x)=-7x+5\to g(x)=-7(x+3)+5 \\ g(x)=-7x-21+5 \\ g(x)=-7x-16 \end{gathered}$

b) The function f stretched vertically by a factor of 3 and translated up by 2 units.

In order to stretch up the function, we multiply the whole function by the factor:

$\begin{gathered} f(x)=7x+5\to f^{\prime}(x)=3\cdot(7x+5) \\ f^{\prime}(x)=21x+15 \end{gathered}$

Then, to translated 2 units up, we add 2 units of the function:

$\begin{gathered} f^{\prime}(x)=21x+15\to g(x)=21x+15+2 \\ g(x)=21x+17 \end{gathered}$

c) The function f translated 2 units down and 3 units right.

To do a translation of 2 units down, we subtract 2 units of the function, and to translate 3 units right, we subtract 3 units from the value of x:

$\begin{gathered} f(x)=7x+5\to f^{\prime}(x)=7x+5-2 \\ f^{\prime}(x)=7x+3 \\ \\ f^{\prime}(x)=7x+3\to g(x)=7\cdot(x-3)+3 \\ g(x)=7x-21+3 \\ g(x)=7x-18 \end{gathered}$

d) The function f stretched vertically by a factor of 2 and translated down by 3 units.

In order to stretch up the function, we multiply the whole function by the factor:

$\begin{gathered} f(x)=7x+5\to f^{\prime}(x)=2\cdot(7x+5) \\ f^{\prime}(x)=14x+10 \end{gathered}$

Then, to translated 3 units down, we subtract 3 units of the function:

$\begin{gathered} f^{\prime}(x)=14x+10\to g(x)=14x+10-3 \\ g(x)=14x+7 \end{gathered}$