A.
B.
C.
D.
Which one is the graph of a function?
- A
- B
- C
- D
Answer.
\(\text{A}\)
Solution.
(A) is the graph of a function. All others fail the vertical line test.
Section 2.6 Exercises
These are the exercises in Module 2.
Checkpoint 2.6.1.
Checkpoint 2.6.2.
If \(f(x) = 3 x^2-x+7\text{,}\) find the following:
(a) \(f(-5)=\)
(b) \(f(2)=\)
(c) \(f(4)=\)
Answer 1.
\(87\)
Answer 2.
\(17\)
Answer 3.
\(51\)
Checkpoint 2.6.3.
Enter Yes or No in each answer space below to indicate whether the corresponding equation defines \(y\) as a function of \(x\text{.}\)
1. \(x^2y+y= 9\)
2. \(4 x = y^2\)
3. \(4 + x = y^3\)
4. \(x + 4 = y^2\)
Answer 1.
\(\text{Yes}\)
Answer 2.
\(\text{No}\)
Answer 3.
\(\text{Yes}\)
Answer 4.
\(\text{No}\)
Checkpoint 2.6.4.
The domain of the function
\(f(x)=\sqrt{x(x-14)}\)
is
Write the answer in interval notation.
Note: If the answer includes more than one interval, write the intervals seperated by the union symbol, U. If needed end \(-\infty\) as -infinity and \(\infty\) as infinity.
Answer.
\(\left(-\infty ,0\right]\cup \left[14,\infty \right)\)
Checkpoint 2.6.5.
One of the solutions of \((5x+2)(x+3)=0\) is:
- \(\displaystyle \frac{5}{2} \)
- \(\displaystyle -\frac{2}{5} \)
- \(\displaystyle \frac{2}{5} \)
- \(\displaystyle 3 \)
- None of the above
Answer.
\(\text{Choice 2}\)
Checkpoint 2.6.6.
Factor: \(x^2+x-30=(\)\()(\))
Note: Your answers should be completely simplified. Unsimplified answers will not be accepted.
Answer 1.
\(x+6\)
Answer 2.
\(x-5\)
Checkpoint 2.6.7.
Find all values of \(x\) that satisfy the equation
\(9(x+8)+2=-8(x-7)-4\)
List the values below, seperated by commas.
Values of \(x\) =
Answer.
\(\frac{-22}{17}\)
Checkpoint 2.6.8.
Find all real zeros of \(f(x)= x^2 - 6 x - 7\text{.}\)
Zeros are \(x =\) .
Note: If there is more than one answer enter them separated by commas.
Answer.
\(-1, 7\)
Checkpoint 2.6.9.
Select the expression that is equivalent to
\(\frac{x-1}{5x}\cdot\frac{x}{x^2-1}\)
- \(\displaystyle \frac{-1}{5(x-1)} \)
- \(\displaystyle \frac{1}{5(x+1)}, x \neq 0, x \neq 1 \)
- \(\displaystyle \frac{-1}{5(x-1)}, x \neq 0, x \neq 1 \)
- \(\displaystyle \frac{1}{5(x+1)} \)
- \(\displaystyle \frac{-1}{5(x-1)}, x \neq 0 \)
- \(\displaystyle \frac{1}{5(x+1)}, x \neq 0 \)
- None of the above
Answer.
\(\text{Choice 2}\)
Checkpoint 2.6.10.
Write \(\frac{m}{2}+\frac{m}{3}\) as a single fraction.
- \(\displaystyle \frac{2m}{6} \)
- \(\displaystyle \frac{m^2}{6} \)
- \(\displaystyle \frac{5m}{5} \)
- \(\displaystyle \frac{5m}{6} \)
- \(\displaystyle \frac{m}{5} \)
- \(\displaystyle \frac{2m}{5} \)
- None of the above
Answer.
\(\text{Choice 4}\)
Checkpoint 2.6.11.
Select the expression that is equivalent to
\(\frac{x-2}{5x}\cdot\frac{x}{x^2-4}\)
- \(\displaystyle \frac{-1}{5(x-2)}, x \neq 0 \)
- \(\displaystyle \frac{1}{5(x+2)}, x \neq 0 \)
- \(\displaystyle \frac{1}{5(x+2)} \)
- \(\displaystyle \frac{1}{5(x+2)}, x \neq 0, x \neq 2 \)
- \(\displaystyle \frac{-1}{5(x-2)}, x \neq 0, x \neq 2 \)
- \(\displaystyle \frac{-1}{5(x-2)} \)
- None of the above
Answer.
\(\text{Choice 4}\)
Checkpoint 2.6.12.
Divide and simplify your answer.
\(\frac{x^{2}+x}{10}\div\frac{x+1}{5}\)
Answer.
\(\frac{x}{2}\)
Checkpoint 2.6.13.
Evaluate the following expression without using a calculator. Simplify your answer as much as possible, and enter your answer as a fraction.
\(\displaystyle \left( \frac{27}{8} \right)^{-1/3}\) =
Answer.
\({\frac{2}{3}}\)
Checkpoint 2.6.14.
Simplify and write the following as a rational power of \(p\) and \(z\text{.}\) If
\begin{equation*}
\displaystyle \frac{ \left(p^5 z^4\right)^{1/15}}
{p^{-1/5}z^{1/3}} = p^m z^n
\end{equation*}
then
\(m =\) and \(n =\)
Answer 1.
\({\frac{8}{15}}\)
Answer 2.
\(-{\frac{1}{15}}\)
Checkpoint 2.6.15.
The expression
\(\left(y^{\frac{3}{6}}\right)^{\frac{2}{5}}\)
equals \(y^r\) where \(r\text{,}\) the exponent of \(y\text{,}\) is:
Answer.
\({\frac{1}{5}}\)
Checkpoint 2.6.16.
The expression
\begin{equation*}
\frac{2^52^42^{-5}}{\sqrt{4^2}2^32^{-5}}
\end{equation*}
equals \(2^n\) where \(n\) is:
Answer.
\(4\)
Solution.
\(\begin{aligned}
\amp \phantom{\mathrel{{}={}}}\frac{2^{5}2^{4}2^{-5} }{ \sqrt{4^{2}}2^{3}2^{-5}} \\
\amp =\frac{2^{5+4-5} }{ ((2^2)^2)^{\frac{1}{2}}2^{3-5}} \\
\amp =\frac{2^{4} }{ 2^{2}2^{-2}} \\
\amp =2^{4 - (2 - 2) } \\
\amp =2^{4}
\end{aligned}\)
Checkpoint 2.6.17.
Consider the polynomial \(p(x)=2x+4x^3 - 10\)
Its degree is and the leading coefficient is
Answer 1.
\(3\)
Answer 2.
\(4\)
Checkpoint 2.6.18.
Determine the following for: \(x^2+ x^6+ x+ 3\)
a) Determine the coefficient and the degree of each term.
| Term | Coefficient | Degree |
| \(x^2\) | ||
| \(x^6\) | ||
| \(x\) | ||
| \(3\) |
b) The degree of the polynomial is ,
the leading term is ,
and the leading coefficient is .
Answer 1.
\(1\)
Answer 2.
\(2\)
Answer 3.
\(1\)
Answer 4.
\(6\)
Answer 5.
\(1\)
Answer 6.
\(1\)
Answer 7.
\(3\)
Answer 8.
\(0\)
Answer 9.
\(6\)
Answer 10.
\(1x^{6}\)
Answer 11.
\(1\)
Checkpoint 2.6.19.
The polynomial \(f(x) = \left(4-x\right)\!\left(x+2\right)\!\left(8x+2\right)^{2}\) has
| degree | \(=\) | |
| leading coefficient | \(=\) | |
| constant coefficient | \(=\) |
Answer 1.
\(4\)
Answer 2.
\(-64\)
Answer 3.
\(32\)
Solution.
SOLUTION
Since \(f(x) = \left(4-x\right)\!\left(x+2\right)\!\left(8x+2\right)^{2} = \left(-x+4\right)\!\left(x+2\right)\!\left(8x+2\right)\!\left(8x+2\right)\text{,}\) by multiplying the highest degree terms in each factor we find the highest degree term to be \((-x)(x)(8 x)(8 x) = -64 x^{4}\text{,}\) and therefore the degree is \(4\) and the leading coefficent is \(-64\text{.}\) Similarly, by multiplying the constant terms from each factor of \(\left(-x+4\right)\!\left(x+2\right)\!\left(8x+2\right)\!\left(8x+2\right)\text{,}\) we get the constant coefficient \((4)(2)(2)(2) = 32\text{.}\) Therefore, \(f(x) = -64 x^{4} + \cdots + 32\text{,}\) where we have omitted the degree 1, 2, and 3 terms.
Checkpoint 2.6.20.
Given that \(f(x)\) is a cubic function with zeros at \(-6\text{,}\) \(-1\text{,}\) and \(7\text{,}\) find an equation for \(f(x)\) given that \(f(-8)=-2\text{.}\)
Answer.
\(\left({\frac{1}{105}}\right)\!\left(x+6\right)\!\left(x+1\right)\!\left(x-7\right)\)
