Functions, Graphs, and Limits - AP Calculus BC

Card 1 of 1344

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

The rate of decrease due to poaching of the elephants in unprotected Sahara is proportional to the population. The population in one region decreased from 1038 to 817 between 2010 and 2015. What is the expected population in 2017?

Tap to reveal answer

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population decreased from 1038 to 817 between 2010 and 2015, we can solve for this constant of proportionality:

$817=1038e^{k(2015-2010)}$

$\frac{817}{1038}=e^{5k}$

$5k=ln(\frac{817}{1038})$

$k=\frac{ln(\frac{817}{1038})}{5}=-0.048$

Using this, we can calculate the expected value from 2015 to 2017:

$P=817e^{(2017-2015)(-0.048)} \approx 742$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the graph above.}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Observation of the graph shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point:}\&x=\frac{2}{17}\&\text{We find a zero denominator for the function: }\&\frac{(13x - 9)}{(17x - 2)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the graph above.}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-\frac{3}{8}\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(6x - 18)}{(16x + 9)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Which of the four following functions is depicted in the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Observation of the graph shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered around:}\&x=1.3\&\text{We find a zero denominator for the function: }\&\frac{(4x + 18)}{(14x - 19)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Which of the four following functions is depicted in the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-0.8\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(5x + 12)}{(6x - 10)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Of the following functions, which most likely represents the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=0.3\&\text{This matches the ratio of coefficients for the function: }\&\frac{(3x - 2)}{(10x - 11)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Of the following functions, which most likely represents the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-1.12\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(19x + 9)}{(17x - 5)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Of the following functions, which most likely represents the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Observation of the graph shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.3\&\text{We find a zero denominator for the function: }\&-\frac{(x - 13)}{(15x + 4)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Of the following functions, which most likely represents the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Observation of the graph shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=0.1\&\text{We find a zero denominator for the function: }\&\frac{(9x - 11)}{(15x - 1)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the graph above.}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=0.71\&\text{This matches the ratio of coefficients for the function: }\&\frac{(12x - 20)}{(17x)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the graph above.}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=1.78\&\text{This matches the ratio of coefficients for the function: }\&\frac{(16x - 13)}{(9x - 4)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Of the following functions, which most likely represents the graph above?}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=0.2\&\text{This matches the ratio of coefficients for the function: }\&\frac{(3x + 14)}{(15x - 16)}\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the graph above.}\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{Notice that as x increases, the function graphed appears to}\&\text{level out, flattening towards a certain value. This value is}\&\text{what is known as a horizontal asymptote. An asymptote is a value}\&\text{that a function approaches, but never actually reaches. Think}\&\text{of a horizontal asymptote as the limit of a function as x approaches}\&\text{infinity. In such a case, as x approaches infinity, any constants}\&\text{added or subtracted in the numerator and denominator become}\&\text{irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-1.67\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(20x + 10)}{(12x - 10)}\end{align}$

← Didn't Know|Knew It →

Question

The graph above is a sketch of the function . For what intervals is continuous?

Tap to reveal answer

Answer

For a function to be continuous at a point , must exist and $\lim_{x \rightarrow a}f(x)=f(a)$ .

This is true for all values of except and .

Therefore, the interval of continuity is $(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$ .

← Didn't Know|Knew It →

Question

Consider the function $f(x)=\frac{x^2-7x+12}{x-3}$ .

Which of the following statements are true about this function?

I. $f(x)\ is\ continuous\ at\ x = 3.$

II. $\lim_{x\to3}f(x)=-1$

III. $\lim_{x\to3}f(x)\ does\ not\ exist$

Tap to reveal answer

Answer

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

$f(3)=\frac{3^2-7(3)+12}{3-3}=\frac{(x-3)(x-4)}{(x-3)}=(x-4)=(3-4)=-1$ .

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

$f(x)=\frac{x^2-7x+12}{x-3}=\frac{(x-4)(x-3)}{(x-3)}=x-4$

$\lim_{x\to3}(x-4)=-1$

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

← Didn't Know|Knew It →

Question

Rewrite as a Cartesian equation:

$x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]$

Tap to reveal answer

Answer

$x = t^{2} + 2t + 1$

$x = \left ( t + 1\right )^{2}$

$\pm \sqrt{x} = t + 1$

So

$\sqrt{x} = t + 1$ or $-\sqrt{x} = t + 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so is nonnegative; we choose

$\sqrt{x} = t + 1$ .

Also,

$y = t^{2} - 2t + 1$

$y= \left ( t - 1\right )^{2}$

$\sqrt{y} = \pm \left ( t - 1\right )$

So

$\sqrt{y} = t- 1$ or $-\sqrt{y} = t- 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so is nonpositive; we choose

$-\sqrt{y} = t- 1$

or equivalently,

$\sqrt{y} = -t+ 1$

to make nonpositive.

Then,

$\sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and

$\sqrt{x}+ \sqrt{y} = 2$

$\sqrt{y} = 2 - \sqrt{x}$

$y =\left ( 2 - \sqrt{x} \right )^2$

$y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$y = x + 4 - 4\sqrt{x}$

← Didn't Know|Knew It →

Question

Rewrite as a Cartesian equation:

$x =10^ {t}, y = \sinh t, t \in (0,\infty )$

Tap to reveal answer

Answer

$y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$ , so

$y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$ .

← Didn't Know|Knew It →

Question

Draw the graph of $r=sin\theta$ from $0\leq \theta \leq2\pi$ .

Tap to reveal answer

Answer

Between and $\frac{\pi }{2}$ , the radius approaches from .

From $\frac{\pi }{2}$ to $\pi$ the radius goes from to .

Between $\pi$ and $\frac{3\pi }{2}$ , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .

From $\frac{3\pi }{2}$ and $2\pi$ , the curve is redrawn in the second quadrant as the radius approaches from .

← Didn't Know|Knew It →

Question

Draw the graph of $r=sin(2\theta )$ where $0\leq \theta \leq2\pi$ .

Tap to reveal answer

Answer

Because this function has a period of $\pi$ , the amplitude of the graph appear at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches 1 from 0.

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches 0 from 1.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches 1 from 0. Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches 0 from 1.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between $\frac{7\pi }{4}$ and $2\pi$ , the curve is drawn in the second quadrant.

← Didn't Know|Knew It →

Question

Graph $r^2=cos(2\theta )$ where $0\leq \theta \leq2\pi$ .

Tap to reveal answer

Answer

Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to $\frac{\pi }{4}$ , $\frac{3\pi }{4}$ to $\frac{5\pi }{4}$ , and $\frac{7\pi }{4}$ to $2\pi$ .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\pm 1$ .

To draw the graph, the radius is 1 at and traces to 0 at $\frac{\pi }{4}$ . As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From $\frac{3\pi }{4}$ to $\pi$ , the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in $\pi$ to $2\pi$ .

← Didn't Know|Knew It →

0

Knew It