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Math Trailhead

Reed College Department of Mathematics & Statistics

Section Math 111

This set of exercises is to help you assess your own algebra background knowledge before beginning Math 111. For more algebra practice resources, as well as links to study and review materials, checkout the complete Reed Math Trailhead.
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Subsection Working with Functions

Evaluating and Interpreting Functions
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Checkpoint 1. Evaluate a Function at Given Inputs.

Evaluate \(f(x)={2x^{2}+6}\) at the following values:
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Answer 1.
\(8\)
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Answer 2.
\(104\)
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Answer 3.
\(2a^{2}+6\)
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Answer 4.
\(8a^{2}+6\)
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Answer 5.
\(2\mathopen{}\left(a+2\right)^{2}+6\)
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Solution.
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\(=2(-1)^2 + 6\)
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\(=2 \cdot 1 + 6\)
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\(= {8}\)
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\(=2(7)^2 + 6\)
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\(=2 \cdot 49 + 6\)
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\(= {104}\)
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\(={2a^{2}+6}\)
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\(2(2a)^2 + 6\)
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\(2\cdot 4a^2 + 6\)
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\({8a^{2}+6}\)
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\(2(a + 2)^2 + 6\)
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\(2({a^{2}+4a+4}) + 6\)
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\({2a^{2}+8a+8} + 6\)
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\({2a^{2}+8a+14}\)
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Checkpoint 2. Evaluate a Piecewise-Defined Function.

Given the piecewise function, find the function values.
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\(f(x)={\begin{cases}\displaystyle{2x-1}\amp \text{if}\ x \lt 0\cr \displaystyle{6-5x}\amp \text{if}\ x \ge 0\end{cases}}\)
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\(f(-3)=\)
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\(f(0)=\)
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\(f(2)=\)
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\(f(8)=\)
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Hint.
First find which part of the piecewise function the desired `x`-value is part of, then use that piece to evaluate the function value.
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Answer 1.
\(-7\)
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Answer 2.
\(6\)
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Answer 3.
\(-4\)
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Answer 4.
\(-34\)
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Composing Functions
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Checkpoint 3. Compose Two Functions.

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Subsection Linear Equations

Equations and Graphs of Lines
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Checkpoint 4. Find a Line Through Two Points.

Find a linear equation satisfying the following conditions:
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Passes through the points \((-1,-1)\) and \((5,-9)\)
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Write your answer using integers or fractions.
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\(y =\)\(x +\)
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Answer 1.
\(-{\frac{4}{3}}\)
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Answer 2.
\(-{\frac{7}{3}}\)
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Solution.
We are given two points:
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\(x_1 = -1\text{,}\) \(y_1 = -1\text{,}\) \(x_2 = 5\text{,}\) and \(y_2 = -9\text{.}\)
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Start by finding the slope:
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\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-9 + 1}{5 + 1} = \frac{{-8}}{{6}} = {-{\frac{4}{3}}}\)
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Now we can use the point-slope formula to write the equation of this line:
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\(y - y_1 = m(x - x_1)\)
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\(m = {-{\frac{4}{3}}}\)
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\(x_1 = -1\)
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\(y_1 = -1\)
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\(y + 1 = {-{\frac{4}{3}}}(x + 1)\)
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To put the equation in slope-intercept form, distribute -4/3 and then add -1 to both sides.
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\({y-\left(-1\right)} = {-{\frac{4}{3}}}x - {-{\frac{4}{3}}} \cdot -1\)
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\({y-\left(-1\right)} = {-\left({\frac{4}{3}}\right)x-{\frac{4}{3}}}\)
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\(y = {-{\frac{4}{3}}}x + {-{\frac{7}{3}}}\)
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Note: To multiply fractions, we multiply straight across. For example \(\frac{2}{5} \cdot 6 = \frac{2}{5} \cdot \frac{6}{1} = \frac{12}{5}\text{.}\)
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To add or subtract fractions we need a least common denominator (LCD). For example, \(\frac{2}{5} - 3 = \frac{2}{5} - \frac{3}{1} = \frac{2}{5} - \frac{15}{5} = -\frac{13}{5}\text{.}\)
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Checkpoint 5. Write a Line from Its Slope and a Point.

Find an equation of the line with slope \(\displaystyle{{{\frac{1}{3}}} }\) that passes through the point \((4,3)\text{.}\) Write your solution in slope-intercept form.
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Answer.
\(y = \left({\frac{1}{3}}\right)x+\left({\frac{5}{3}}\right)\)
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Solution.
When trying to find the equation of a line with some information, start with the equation in slope-intercept form, that is, \(y = mx+b\text{.}\) Then substitute in any known information. In this case, we know the slope is \(m={{\frac{1}{3}}}\text{,}\) or
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\(\displaystyle{ y = {{\frac{1}{3}}} x + b }\)
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Next, substitute in the point \((4,3)\) or
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\begin{equation*} \begin{aligned} 3 \amp = ({{\frac{1}{3}}})(4) + b \\ 3 \amp = {{\frac{4}{3}}} + b \end{aligned} \end{equation*}
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Next, substract \({{\frac{4}{3}}}\) from both sides:
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\begin{equation*} {{\frac{5}{3}}} = b \end{equation*}
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and finally plug this value of b into the equation:
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\begin{equation*} y = {{\frac{1}{3}}} x + {{\frac{5}{3}}} \end{equation*}
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Checkpoint 6. Read and Model a Line from Its Graph.

The graph of the function \(y=f(x)\) is given by the line displayed above.
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Find \(f(2)=\)
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Find \(f(5)=\)
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So, the slope is \(m=\)
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Find an equation for the line graphed above:
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  • Do not use decimal approximations in your answer.
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Hint.
Start by identifying a couple of “nice” points on the grid that the line passes through.
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Once have two points for your line to pass through, you can determine the slope.
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Use the point-slope form of a line. \(y = m (x-x_A) + y_A\)
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Answer 1.
\(1\)
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Answer 2.
\(3\)
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Answer 3.
\({\frac{2}{3}}\)
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Answer 4.
\(y = \frac{2}{3}\mathopen{}\left(x-2\right)+1\)
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Solution.
The graphed line passes through the points: \(A = (-1,-1)\text{,}\) \(B = (2,1)\text{,}\) \(C = (5,3)\)
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Select any pair of points on the line to compute the slope:
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\(m = \frac{\Delta y}{\Delta x} = \frac{1 + 1}{2 + 1} = {{\frac{2}{3}}}\)
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Use the point-slope form of a line:
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\(y = m (x - x_A) + y_A\)
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\(y = {{\frac{2}{3}}\mathopen{}\left(x-2\right)+1}\)
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Subsection Polynomial and Rational Expressions

Simplifying Polynomials
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Checkpoint 7. Expand and Simplify a Polynomial Expression.

Part 1 of 2:
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Expand and simplify \(\displaystyle h(t) = t^2 - t(t-5)\text{.}\)
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\(h(t)\) =
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Answer.
\(5t\)
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Factoring Polynomials
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Checkpoint 8. Factor a Quadratic Polynomial.

Factor the following polynomial expression.
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\(\displaystyle{ {12x^{2}-5x-3} }\)
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Factored Form:
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Answer.
\(\left(3x+1\right)\mathopen{}\left(4x-3\right)\)
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Zeros and Intercepts
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Checkpoint 9. Find the Intercepts of a Polynomial.

Find the intercepts of \(f(x)={5\mathopen{}\left(x-5\right)\mathopen{}\left(x-2\right)\mathopen{}\left(x-3\right)}\text{.}\) Enter intercepts as points. If you have more than one point, enter them as a comma separated list.
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Answer 1.
\(\left(0,-150\right)\)
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Answer 2.
\(\left(5,0\right), \left(2,0\right), \left(3,0\right)\)
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Rational Equations and Functions
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Checkpoint 10. Solve a Rational Equation.

Solve the rational equation. If there is more than one correct answer, enter a comma separated list. If there is no answer, enter “NONE”.
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1+\(\dfrac{8}{p}=-\dfrac{12}{p^2}\)
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\(p\) =
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Answer.
\(-2, -6\)
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Checkpoint 11. Find a Removable Discontinuity.

Let \(\displaystyle{ f(x)={\frac{x^{2}-36}{x+6}} }\text{.}\) Identify the removable discontinuity.
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  • Removable Discontinuity: \(x=\)
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Answer.
\(-6\)
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Subsection Exponents and Logarithms

Exponent Rules
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Checkpoint 12. Apply the Power and Product Rules.

Checkpoint 13. Simplify an Exponential Expression.

Logarithm Basics
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Checkpoint 14. Evaluate Basic Logarithms.

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Subsection Trigonometry

Exact Values and Right Triangles
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Checkpoint 15. Find Exact Cosine Values on the Unit Circle.

Without using a calculator, state the exact value of the following trig functions for the specified angle.
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  1. \(\displaystyle{ \cos\left({\frac{\pi }{4}}\right) = }\)
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  2. \(\displaystyle{ \cos\left({\frac{3\pi }{4}}\right) = }\)
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  3. \(\displaystyle{ \cos\left({\frac{5\pi }{4}}\right) = }\)
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  4. \(\displaystyle{ \cos\left({\frac{7\pi }{4}}\right) = }\)
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  5. \(\displaystyle{ \cos\left({\frac{9\pi }{4}}\right) = }\)
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  6. \(\displaystyle{ \cos\left({\frac{-3\pi }{4}}\right) = }\)
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  7. \(\displaystyle{ \cos\left({\frac{-5\pi }{4}}\right) = }\)
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  8. \(\displaystyle{ \cos\left({\frac{-7\pi }{4}}\right) = }\)
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Hint.
Do you notice anything about all the angles in this problem?
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What is the reference angle for each?
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And what does that tell you about the cosine of each angle?
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Answer 1.
\(\frac{\sqrt{2}}{2}\)
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Answer 2.
\(-\frac{\sqrt{2}}{2}\)
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Answer 3.
\(-\frac{\sqrt{2}}{2}\)
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Answer 4.
\(\frac{\sqrt{2}}{2}\)
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Answer 5.
\(\frac{\sqrt{2}}{2}\)
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Answer 6.
\(-\frac{\sqrt{2}}{2}\)
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Answer 7.
\(-\frac{\sqrt{2}}{2}\)
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Answer 8.
\(\frac{\sqrt{2}}{2}\)
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Solution.
Right off the bat, you should recognize that all angles appearing in this problem have \({\frac{\pi }{4}}\) as their reference angle.
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This means that the only difference in your answers will be whether your answer will be positive or negative.
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The answer, of course, depends on the quadrant where \(\theta\) lives.
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From our unit circle, we know that \(\cos({\frac{\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
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\({\frac{\pi }{4}}\) corresponds to the point \(({\frac{\sqrt{2}}{2}},{\frac{\sqrt{2}}{2}})\) on the unit circle;
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and \(\cos(\theta)\) corresponds to the \(x\)-coordinate.
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a. This is our reference angle. \({\frac{\pi }{4}}\) is in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
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b. \({\frac{3\pi }{4}}\) is in the second quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{3\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
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c. \({\frac{5\pi }{4}}\) is in the third quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{5\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
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d. \({\frac{7\pi }{4}}\) is in the fourth quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{7\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
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e. \({\frac{9\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{9\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
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f. \({\frac{-3\pi }{4}}\) is coterminal with \({\frac{5\pi }{4}}\text{,}\) and both are in the third quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{-3\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
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g. \({\frac{-5\pi }{4}}\) is coterminal with \({\frac{3\pi }{4}}\text{,}\) and both are in the second quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{-5\pi }{4}}) = {-\frac{\sqrt{2}}{2}}\text{.}\)
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h. \({\frac{-7\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{-7\pi }{4}}) = {\frac{\sqrt{2}}{2}}\text{.}\)
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Checkpoint 16. Solve a Right Triangle.

Refer to the right triangle in the figure. Click on the picture to see it more clearly.
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If , \(BC=8\) and the angle \(\alpha=50 ^\circ\text{,}\) find any missing angles or sides. Give your answer to at least 3 decimal digits.
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AB =
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AC =
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\(\beta\)=
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Answer 1.
\(6.7127970630123\)
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Answer 2.
\(10.4432583233963\)
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Answer 3.
\(40\)
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