College Algebra - Transformations

Which of these parabolas has its vertex at (5,1)?

None of the other answers.

Explanation

The correct answer is $\boldsymbol{(x-5)^2+1}$ . Inside the portion being squared the distance moved is opposite the sign and is horizontal. Outside the squared portion the distance moved follows the sign (plus is up and minus is down) and is vertical.

For example the incorrect answer would have its vertex at (1,-5).

If we want a function to be reflected about the origin, what would the corresponding equation look like?

Explanation

To compute a reflection about the x-axis, calculate , and to calculate a reflection about the y-axis, calculate . To compute a reflection about the origin, simply combine both reflections into .

In our case, .

So,

The graph of a function is reflected about the -axis, then translated upward $\frac{8}{3}$ units. Which of the following is represented by the resulting graph?

$g(x)= -f(x) + \frac{8}{3}$

$g(x)= f(-x) + \frac{8}{3}$

$g(x)= -f\left (x + \frac{8}{3} \right )$

$g(x)= -f\left (-x + \frac{8}{3} \right )$

$g(x)= -f\left (x- \frac{8}{3} \right )$

Explanation

Reflecting the graph of a function about the -axis results in the graph of the function

$f_{1}(x)= -f(x)$ .

Translating this graph upward $\frac{8}{3}$ results in the graph of the function

$g(x)= f_{1}(x) + \frac{8}{3 }= -f(x)+ \frac{8}{3 }$ .

Which of the following represents a horizontal transformation of v(t) 3 units to the right?

$\frac{1}{3}v(t)$

Explanation

Which of the following represents a horizontal transformation of v(t) 3 units to the right?

To perform a horizontal transformation on a function, we need to add or subtract a value within the function, which looks something like this:

$v(t\pm n)$

Now, counter intuitively, when we shift right, we will subtract. If we wanted to shift left, we would add.

So, to shift 3 to the right, we need:

What is the expression for this polynomial:

after being shifted to the right by 2?

Explanation

To shift a polynomial to the right by 2, we must replace x with x-2 in whatever the expression for the polynomial is. The logic of this is that every x value has a y value associated with it, and we want to give every x value the y value associated with the point that is 2 before it.

So, to get our shifted polynomial, we plug in x-2 as noted.

$x^2+4x-7 \rightarrow (x-2)^2+4(x-2)-7$

and then we combine like terms:

Give the equation of the vertical asymptote of the graph of the equation

$f(x) =4 \log (x- 3)- 2$ .

Explanation

Let $g(x) = \log x$ . In terms of ,

The graph of has as its vertical asymptote the line of the equation . The graph of is the result of three transformations on the graph of - a right shift of 3 units ( ), a vertical stretch ( ), and a downward shift of 2 units ( ). Of the three transformations, only the right shift affects the position of the vertical asymptote; the asymptote of also shifts right 3 units, to .

The graph of a function is shown below, select the graph of

Problem 8 correct

Problem 6 correct transformation

Problem 8 transformation wrong 2

Problem 8 transformation wrong 1

Problem 8 transformation wrong 4

Explanation

There are four fundamental transformations that allows us to think of a function as a transformation of a function ,

In our case, and , so the width and/or height of our function will not change in the coordinate plane.

We have and . The number will shift the function up units along the -axis on the coordinate plane. The number will shift unit to the right on the coordinate plane.

Problem 6 correct transformation

Reflect the graph of $f(x) = \ln x$ about the -axis to yield the graph of a function . Which of the following is a valid way of stating the definition of ?

$g(x)= \ln (-x)$

$g(x)= \ln \frac{1}{x}$

$g(x) = \frac{1}{\ln x}$

$g(x) =e^{x}$

None of the other choices gives the correct response.

Explanation

The reflection of the graph of a function about the -axis yields the graph of the function . Therefore, set $f(x) = \ln x$ and substitute for to yield the function

$g(x) = \ln (-x)$ .

Consider an exponential function . If we want to reflect this function across the y-axis, which of the following equations would result in the desired reflection?

$y=e^{-x}$

$y=-e^{-x}$

Explanation

As a general rule, if you have a function , then in order to reflect across the x-axis, we compute , and in order to reflect across the y-axis, we compute . In our case, we are asked to compute the latter.

So, if , then $f(-x)=e^{-x}$ .

Translate the graph of $f(x) = \ln x$ upward three units to yield the graph of a function . Which of the following is a valid way of stating the definition of ?

$g(x)= \ln (e^{3} x)$

$g(x)= \ln x^{3}$

$g(x)= \ln \left ( \frac{x}{e^{3}} \right )$

$g(x)= \ln (x-3)$

$g(x)= \ln (x+3)$

Explanation

A vertical translation of the graph of a function by units yields the graph of the function . A translation in an upward direction is a positive translation, so setting $f(x)= \ln x$ and , the resulting graph becomes