Graphing

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True or false: The graph of $f(x) = \frac{ x-7}{2x^{2}-15x+7}$ has as a vertical asymptote the graph of the equation .

False

True

Explanation

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator. It is a quadratic trinomial with lead term $2x^{2}$ , so look to factor

$2x^{2}-15x+7$

by using the grouping technique. We try finding two integers whose sum is and whose product is ; with some trial and error we find that these are and , so:

$2x^{2}-15x+7$

Break the linear term:

$=2x^{2}-x-14 x+7$

Regroup:

$=(2x^{2}-x )-(14 x-7)$

Factor the GCF twice:

Therefore, can be rewritten as

$f(x) = \frac{ x-7}{(x -7) (2 x-1)}$

Cancel the common factor from both halves; the function can be rewritten as

$f(x) = \frac{1}{ 2 x-1}$

Set the denominator equal to 0 and solve for :

$\frac{2x}{2}= \frac{1}{2}$

$x= \frac{1}{2}$

The graph of therefore has one vertical asymptotes, the line of the equations $x= \frac{1}{2}$ . The line of the equation is not a vertical asymptote.

Find the point that corresponds to the following ordered pair:

Explanation

In order to get to the point , start at the origin and move right unit and up units.

This is point .

The points and are plotted on a quadrant. Which graph depicts the correct points?

Explanation

A quadrant is set up in such a way that the positive numbers are to the right and top, while the negative numbers are to the left and bottom. The x-value (the first number in the ordered pair) is the distance left or right from the center. The y-value (the second number in the ordered pair) is the distance above or below the axis.

will be units to the right, and units up.

will be units to the left, and units up.

Define a function as follows:

$f(x) = 2 ^{x-3}+ 5$

Give the vertical aysmptote of the graph of .

The graph of does not have a vertical asymptote.

$x= - \frac{5}{2}$

Explanation

Since any number, positive or negative, can appear as an exponent, the domain of the function $f(x) = 2 ^{x-3}+ 5$ is the set of all real numbers; in other words, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

Define a function as follows:

$f(x) = 2 ^{x-1} + 7$

Give the horizontal aysmptote of the graph of .

$y = -\frac{7}{2}$

Explanation

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, $2 ^{x-1} > 0$ and $y= 2 ^{x-1} + 7 > 7$ for all real values of . The graph will never crosst the line of the equatin , so this is the horizontal asymptote.

The length of line segment $\overline{AB}$ is 12 units. If point A is located at , what is a possible location for point B?

$(14, 5+2\sqrt{11})$

$(2\sqrt{11}, 10)$

$(4+2\sqrt{22}, 15)$

Explanation

To answer this question, we will have to manipulate the distance formula:

$d=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}$

To get rid of the square root, we can square both sides:

$d^{2}=(x_{1}-x_{2})^{2}}+(y_{1}-y_{2})^{2}}$

and plug in the information given in the question.

$12^{2}=(4-x_{2})^{2}}+(5-y_{2})^{2}}$

$144=(4-x_{2})^{2}}+(5-y_{2})^{2}}$

At this point we can simply plug in the possible values to determine which combination of coordinates will make the equation above true.

$144=(4-14)^{2}}+(5-(5+2{\sqrt{11}}))^{2}}$ $=(-10)^{2}+(2\sqrt{11})^{2}=100+44=144$

Thus the correct coordinate is,

$(14, 5+2\sqrt{11})$ .

Give the domain of the function

$f(x) =x+ \sqrt{x+ 7}$

$\left \{ x| x \ge -7 \right \}$

The set of all real numbers

$\left \{ x| x \ne -7 \right \}$

$\left \{ x| x \ne 7 \right \}$

$\left \{ x| x \ge 7 \right \}$

Explanation

The square root of a real number is defined only for nonnegative radicands; therefore, the domain of is exactly those values for which the radicand is nonnegative. Solve the inequality:

$x+ 7 \ge 0$

$x+ 7 - 7 \ge 0 - 7$

$x \ge - 7$

The domain of is $\left \{ x| x \ge -7 \right \}$ .

What is the domain of $y = 4 - x^{2}$ ?

all real numbers

$x \leq 4$

$x \geq 4$

$x \leq 0$

Explanation

The domain of the function specifies the values that can take. Here, $4-x^{2}$ is defined for every value of , so the domain is all real numbers.

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

$(-\infty , 27)\cup (27, \infty )$

$(27, \infty )$

$(3, \infty )$

$(-\infty , - 27)\cup (27, \infty )$

$(-\infty , - 3)\cup (3, \infty )$

Explanation

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which