Albertson AP Calculus AB Name___________________________

AP CALCULUS AB SUMMER PACKET 2015

DUE DATE: The beginning of class on the last class day of the first week of school. This assignment is to be done at you leisure during the summer. It is meant to help you practice mathematical skills necessary to be successful in Calculus AB. All of the skills covered in this packet are skills from Algebra 2 and Pre-Calculus. If you need to use reference materials please do so. While graphing calculators will be used in class the majority of this packet should be done without one. If it says to you use one then please do otherwise please refrain. As you know AP Calculus AB is a fast paced course that is taught at the college level. There is a lot of material in the curriculum that must be covered before the AP exam in May. The better you know the prerequisite skills coming into the class the better the class will go for you. Spend some time with this packet and make sure you are clear on everything covered. If you have questions please contact me via email and I will be glad to help. (If you take a picture of your work and the questions it usually makes things go faster) This assignment will be collected and graded as your first test. Be sure to show all appropriate work. In addition, there may be a quiz on this material during the first quarter. All questions must be complete with the correct work. I am very excited to see you all in August. Good Luck!!

I. Intercepts The x-intercept is where the graph crosses the x-axis. You can find the x-intercept by setting 𝑦 = 0. The y-intercept is where the graph crosses the y-intercept. You can find the y-intercept by setting 𝑥 = 0 Example: Find the intercepts for 𝑦 = (𝑥 + 3)2 − 4 Solution X-intercept Set 𝑦 = 0.

0 = (𝑥 + 3)2 − 4 Add 4 to both sides.

4 = (𝑥 + 3)2 Take the square root of both sides

±2 = (𝑥 + 3) Write as two equations

−2 = (𝑥 + 3) 𝑜𝑟 2 = (𝑥 + 3) Subtract 3 from both sides

−5 = 𝑥 𝑜𝑟 − 1 = 𝑥 Y-intercept Set 𝑥 = 0

𝑦 = (0 + 3)2 − 4 Add 0 + 3

𝑦 = 32 − 4 Square 3

𝑦 = 9 − 4 Add four to both sides

𝑦 = 5

Find the intercepts for each of the following.

1. 𝑦 = −3𝑥 + 2

2. 𝑦 = 𝑥3 + 2

3. 𝑦 = 𝑥2+3𝑥(3𝑥+1)2

4. 𝑦2 = 𝑥3 − 4𝑥

II. Complex Fractions When simplifying fractions, multiply by a fraction equal to 1 which has a numerator and denominator composed of the common denominator of all the denominators in the complex fraction. Example

−7− 6𝑥 + 1

5𝑥 + 1

=−7− 6

𝑥 + 15

𝑥 + 1∙𝑥 + 1𝑥 + 1

=−7𝑥 − 7 − 6

5

=−7𝑥 − 13

5

−2𝑥 + 3𝑥

𝑥 − 45 − 1

𝑥 − 4=−2𝑥 + 3𝑥

𝑥 − 45 − 1

𝑥 − 4∙𝑥(𝑥 − 4)𝑥(𝑥 − 4)

=−2(𝑥 − 4) + 3𝑥(𝑥)5(𝑥)(𝑥 − 4) − 1(𝑥)

=−2𝑥 + 8 + 3𝑥2

5𝑥2 − 20𝑥 − 𝑥

=3𝑥2 − 2𝑥 + 8

5𝑥2 − 21𝑥

1. 25𝑎 −𝑎

5+𝑎

2. 2− 4

𝑥+2

5+ 10𝑥+2

3. 4− 12

2𝑥−3

5+ 152𝑥−3

III. System of Equations Use substitution or elimination method to solve the system of equations. Example:

𝑥2 + 𝑦2 − 16𝑥 + 39 = 0 𝑥2 − 𝑦2 − 9 = 0

Elimination Method Add the two equations and you get

2𝑥2 − 16𝑥 + 30 = 0 𝑥2 − 8𝑥 + 15 = 0

(𝑥 − 3)(𝑥 − 5) = 0 𝑥 = 3 𝑎𝑛𝑑 𝑥 = 5

Plug 𝑥 = 3 and 𝑥 = 5 into an original equation.

32 − 𝑦2 − 9 = 0 −𝑦2 = 0 𝑦 = 0

52 − 𝑦2 − 9 = 0

16 = 𝑦2 𝑦 = ±4

Points of intersection are (5,4), (5,−4),𝑎𝑛𝑑 (3,0) Substitution Solve one equation for a variable

𝑦2 = −𝑥2 + 16𝑥 − 39 Plug 𝑦2 into the other equation

𝑥2 − (−𝑥2 + 16𝑥 − 39) − 9 = 0

2𝑥2 − 16𝑥 + 30 = 0 𝑥2 − 8𝑥 + 15 = 0

(𝑥 − 3)(𝑥 − 5) = 0 The rest is like the previous example

Find the point(s) of intersection of the graphs for the given equations.

1. 𝑥 + 𝑦 = 8, 4𝑥 − 𝑦 = 7

2. 𝑥2 + 𝑦 = 6, 𝑥 + 𝑦 = 4

3. 𝑥2 − 4𝑦2 − 20𝑥 − 64𝑦 − 172 =0, 16𝑥2 + 4𝑦2 − 320𝑥 + 64𝑦 +1600 = 0

IV. Functions

Let 𝑓(𝑥) = 2𝑥 + 1 𝑎𝑛𝑑 𝑔(𝑥) = 2𝑥2 − 1. Find each

To evaluate a function for a given value, simply plug the value into the function for x. (𝑓°𝑔)(𝑥) = 𝑓�𝑔(𝑥)� 𝑂𝑅 𝑓[𝑔(𝑥)] read “f of g of x”. Means to plug the inside function (in this case 𝑔(𝑥) in for x in the outside function (in this case, 𝑓(𝑥)). Example Given 𝑓(𝑥) = 2𝑥2 + 1 and 𝑔(𝑥) = 𝑥 − 4 find 𝑓�𝑔(𝑥)�

𝑓�𝑔(𝑥)� = 𝑓(𝑥 − 4) = 2(𝑥 − 4)2 + 1= 2(𝑥2 − 8𝑥 + 16) + 1= 2𝑥2 − 16𝑥 + 32 + 1

𝑓�𝑔(𝑥)� = 2𝑥2 − 16𝑥 + 33

1. 𝑓(2) =

2. 𝑔(−3) =

3. 𝑓(𝑡 + 1)

4. 𝑓�𝑔(−2)� =

5. 𝑔�𝑓(𝑚 + 2)� =

Find 𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ for the given

function f 6. 𝑓(𝑥) = 9𝑥 + 3

7. 𝑓(𝑥) = 5 − 2𝑥

V. Interval Notation Solve each equation. State your

Complete the table with the appropriate notation or graph

Solution Interval Notation

Graph

−2 < 𝑥 < 4

[−1,7)

answer in BOTH interval notation and graphically.

1. 2𝑥 − 1 ≥ 0

2. −4 ≤ 2𝑥 − 3 < 4

3. 𝑥2− 𝑥

3> 5

VI. Domain and Range The domain of a function is the set of x values for which the function is defined. The range of a function is the set of y values that a function can return. In Calculus we usually write domains and ranges in interval notation. Example: Find the domain and range for 𝑓(𝑥) = √𝑥 − 3 Solution Since we can only take the square root of positive numbers 𝑥 − 3 ≥ 0 which means that 𝑥 ≥ 3. So we would say the domain is [3,∞). Note that we have used a [ to indicate that 2 is included. If 3 was not to be included we would have used a (. The smallest y value that the function can return is 0 so the range is [0,∞)

Find the domain and range.

1. ℎ(𝑥) = √9 − 𝑥2

2. ℎ(𝑥) = sin𝑥

3. 𝑓(𝑥) = 2𝑥−1

VII. Inverses To find the inverse of a function, simply switch the x and the y and solve for the new “y” value.

𝑓(𝑥) = √𝑥 + 13 Rewrite 𝑓(𝑥) as y

𝑦 = √𝑥 + 13 Switch x and y

𝑥 = �𝑦 + 13 Solve for your new y.

𝑥3 = �𝑦 + 13 3

𝑥3 = 𝑦 + 1 𝑦 = 𝑥3 − 1

Rewrite in inverse notation

𝑓−1(𝑥) = 𝑥3 − 1 To prove that one function is an inverse of another function, you need to show that 𝑓�𝑔(𝑥)� = 𝑔�𝑓(𝑥)� =𝑥

Find the inverse of each function

1. 𝑓(𝑥) = 2𝑥 + 1

2. 𝑓(𝑥) = 𝑥3

3

Prove that f and g are inverse of each other

3. 𝑓(𝑥) = 𝑥3

2

𝑔(𝑥) = √2𝑥3

4. 𝑓(𝑥) = 9 − 𝑥2,𝑥 ≥ 0 𝑔(𝑥) = √9 − 𝑥

VIII. Symmetry x-axis substitute in –y for y into the equation. If this yields an equivalent equation then the graph has x-axis symmetry. If this is the case, this is not a function as it would fail the vertical line test. y-axis substitute in –x for x into the equation. If this yields an equivalent equation then the graph has y-axis symmetry. A function that has y-axis symmetry is called an even function. Origin Substitute in -x for x into the equation and substitute –y for y into the equation. If this yields an equivalent equation then the graph has origin symmetry. If a function has origin symmetry it is called an odd function. In order for a graph to represent a function it must be true that for every x value in the domain there is exactly one y value. To test to see if an equation is a function we can graph it and then do the vertical line test. Example 1 Is 𝑥 − 𝑦2 = 2 a function? Solution: this is not a function because it does not pass the vertical line test.

Example 2: Test for symmetry with respect to each axis and the origin given the equation 𝑥𝑦 − √4 − 𝑥2 = 0

Test for symmetry with respect to each axis and the origin.

1. 𝑦 = 𝑥√𝑥 + 2

2. 𝑦 = |6 − 𝑥|

3. 𝑦 = 𝑥𝑥3+1

Solution: x-axis

𝑥(−𝑦) −�4 − 𝑥2 = 0 −𝑥𝑦 − √4 − 𝑥2 = 0 since there is no way to make this look like the original it is not symmetric to the x axis y-axis

−𝑥𝑦 − �4 − (−𝑥)2 = 0 −𝑥𝑦 − √4 − 𝑥2 = 0 since there is no way to make this look like the original it is not symmetric to the y axis Origin

−𝑥(−𝑦) −�4 − (−𝑥)2 = 0 𝑥𝑦 − √4 − 𝑥2 = 0 since this does look like the original it is symmetric to the origin

4. 3𝑥2 − 1

IX. Vertical Asymptotes To find the vertical asymptotes, set the denominator equal to zero to find the x-value for which the function is undefined. That will be the vertical asymptote.

Determine the vertical asymptotes for the function (it will be a line)

1. 𝑓(𝑥) = 1𝑥2

2. 𝑓(𝑥) = 𝑥2

𝑥2−4

3. 𝑓(𝑥) = 2+𝑥𝑥2(1−𝑥)

X. Holes (Points of Discontinuity) Given a rational function if a number causes the denominator and the numerator to be 0 then both the numerator and denominator can be factored and the common zero can be cancelled out. This means there is a hole in the function at this point. Example 1 Find the hole in the following function

𝑓(𝑥) =𝑥 − 2

𝑥2 − 𝑥 − 2

Solution: When 𝑥 = 2 is substituted into the function the denominator and numerator both are 0. Factoring and cancelling 𝑓(𝑥) = 𝑥−2

(𝑥+1)(𝑥−2)

𝑓(𝑥) = 1𝑥+1

but 𝑥 ≠ 2 this restriction is from the original function before canceling. The graph of the function 𝑓(𝑥) witll look identical to 𝑓(𝑥) = 1

𝑥+1 except

for the hole at 𝑥 = 2

For each function list find the holes

1. 𝑓(𝑥) = (𝑥−3)(𝑥+2)(𝑥−3)(2𝑥+1)

2. 𝑓(𝑥) = 𝑥2−12𝑥2+𝑥−1

XI. Horizontal Asymptotes Case 1: Degree of the numerator is less than the degree of the denominator. The asymptote is 𝑦 = 0 Case 2: Degree of the numerator is the same as the degree of the denominator. The asymptote is the ratio of the lead coefficients. Case 3: Degree of the numerator is greater than the degree of the denominator. There is no horizontal asymptote. The function increases without bound. (If the degree of the numerator is exactly 1 more than the degree of the denominator, then there is a slant asymptote, which is determined with long division.)

Determine the horizontal asymptotes using the three cases.

1. 𝑓(𝑥) = 𝑥2−2𝑥+1𝑥3+𝑥−7

2. 𝑓(𝑥) = 3𝑥3−2𝑥2+84𝑥−3𝑥3+5

3. 𝑓(𝑥) = 4𝑥5

𝑥2−7

XII. Solving for indicated variables

1. 𝑥𝑎

+ 𝑦𝑏

+ 𝑧𝑐

= 1 𝑓𝑜𝑟 𝑎

2. 𝑉 = 2(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎)𝑓𝑜𝑟 𝑎

3. 2𝑥 − 2𝑦𝑑 = 𝑦 + 𝑥𝑑 𝑓𝑜𝑟 𝑑

4. 2𝑥4𝜋

+ 1−𝑥2

= 0 for x

XIII. Absolute Value and Piecewise Functions In order to remove the absolute value sign from a function you must:

1. Find the zeros of the expression inside of the absolute value.

2. Make a sign chart of the expression inside the absolute value

3. Rewrite the equation without the absolute value as a piecewise function. For each interval where the expression is positive we can write that interval by just dropping the absolute value. For each interval that is negative we must take the opposite sign.

Example 1 Rewrite the following equation without using absolute value symbols.

𝑓(𝑥) = |2𝑥 + 4| Solution:

Write the following absolute value expressions as piecewise expressions (by remove the absolute value):

1. 𝑦 = |2𝑥 − 4|

Find where the expression is 0 for the part in the absolute value

2𝑥 + 4 = 0 2𝑥 = −4

𝑥 = −42

𝑥 = −2 Put in any value less than -2 into 2x+4 and you get a negative. Put in any value more than -2 and you get a positive.

Write as a piecewise function. Be sure to change the sign of each term for any part of the graph that was negative on the sign chart.

𝑓(𝑥) = �−2𝑥 − 4 𝑥 < −22𝑥 + 4 𝑥 ≥ −2

�

2. 𝑦 = |6 + 2𝑥| + 1

XIV. Exponents A fractional exponent means you are taking a root.

For example 𝑥12 is the same as √𝑥

Example 1:

Write without fractional exponent: 𝑦 = 𝑥23

Solution: 𝑦 = √𝑥23 Notice that the index is the denominator and the power is the numerator. Negative exponents mean that you need to take the reciprocal. For example 𝑥−2 means 1

𝑥2 and 2

𝑥−3

means 2𝑥3. Example 2: Write with positive exponents: 𝑦 = 2

5𝑥−4

Solution: 𝑦 = 2𝑥4

5

Example 3: Write with positive exponents and

Write without fractional exponents

1. 𝑦 = 2𝑥13

2. 𝑓(𝑥) = (16𝑥2)14

3. 𝑦 = 2713𝑥

34

4. 912 =

5. 6413

6. 823 =

without fractional exponents: 𝑓(𝑥) = (𝑥+1)−2(𝑥−3)12

(2𝑥−3)−12

Solution: 𝑓(𝑥) = √𝑥−3√2𝑥−3(𝑥+1)2

When factoring always factor out the lowest exponent for each term. Example 4: 𝑦 = 3𝑥−2 + 6𝑥 − 33𝑥−1 Solution: 𝑦 = 3𝑥−2(1 + 2𝑥3 − 11𝑥) When dividing two terms with the same base, we subtract the exponents. If the difference is negative then the term goes in the denominator if the difference is positive then the term goes in the numerator.

Example 5: Simplify 𝑓(𝑥) = (2𝑥)3

𝑥8

Solution: first you must distribute the exponent.

𝑓(𝑥) = 8𝑥3

𝑥8. Then since we have two terms with x as

the base we can subtract the exponents. Thus 𝑓(𝑥) = 8

𝑥5

Example 6: Factor and simplify

𝑓(𝑥) = 4𝑥(𝑥 − 3)12 + 𝑥2(𝑥 − 3)−

12

Solution: The common terms are x and (x-3). The lowest exponent for x is 1. The lowest exponents for

(x-3) is −12. So factor out 𝑥(𝑥 − 3)−

12 and obtain

𝑓(𝑥) = 𝑥(𝑥 − 3)−12[4(𝑥 − 3) + 𝑥]

This will simplify to

𝑓(𝑥) = 𝑥(𝑥 − 3)−12[4𝑥 − 12 + 𝑥]

Leaving a final solution of 𝑥(5𝑥−12)√𝑥−3

Write with positive exponents:

7. 𝑓(𝑥) = 2𝑥−3

8. 𝑦 = � −2𝑥−4

�−2

Factor then simplify

9. 𝑓(𝑥) = 4𝑥−3 + 2𝑥 − 18𝑥−2

10. 5𝑥2(𝑥 − 2)−12 + (𝑥 − 2)

123𝑥 = 𝑦

11. 𝑓(𝑥) = 6𝑥(2𝑥 − 1)−1 − 4(2𝑥 −1)

XV. Natural Logarithms Recall that 𝑦 = ln (𝑥) and 𝑦 = 𝑒𝑥 are inverse to each other. Properties of Natural Log:

ln(𝐴𝐵) = ln𝐴 + ln𝐵 Example 1: ln(2) + ln(5) = ln (10)

ln �𝐴𝐵� = ln𝐴 − ln𝐵

Example 2: ln 6 − ln 2 = ln 3

ln𝐴𝑝 = 𝑝 ln𝐴 Example 3: ln𝑥4 = 4 ln 𝑥

3ln 2 = ln 23 = ln 8

ln(𝑒𝑥) = 𝑥 , ln 𝑒 = 1, ln 1 = 0, 𝑒0 = 1 Example 4: Use the properties of natural logs to solve for x.

2 ∙ 5𝑥 = 11 ∙ 7𝑥 5𝑥

7𝑥=

112

ln5𝑥

7𝑥= ln

112

ln 5𝑥 − ln 7𝑥 = ln 11 − ln 2 𝑥𝑙𝑛 5 − 𝑥 ln 7 = ln 11 − ln 2 𝑥(ln 5 − ln 7) = ln 11 − ln 2

𝑥 =ln 11 − ln 2ln 5 − ln 7

Express as a single logarithm:

1. 3 ln𝑥 + 2 ln𝑦 − 4 ln 𝑧 Solve for x

2. 3 ln𝑥 = 1

3. 𝑒𝑥−3 = 7

4. 3𝑥 = 5 ∙ 2𝑥

XVI. Radians and Degree Measure Use 180°

𝜋𝑟𝑎𝑑𝑖𝑎𝑛𝑠 to get rid of radians and convert to

degrees Use 𝜋𝑟𝑎𝑑𝑖𝑎𝑛𝑠

180° to get rid of degrees and convert to

radians

Convert to degrees

1. 5𝜋6

2. 2.63 radians

Convert to radians

3. 45°

4. -17°

5. 237° XVII. Trig. Equations and special values

You are expected to know the special values for trigonometric functions. Fill in the table to the right and study it. (Please) You can determine sine or cosine of a quadrantal angle by using the unit circle. The x-coordinate is the cosine and the y-coordinate is the sine of the angle.

Example:

sin 90° = 1

cos𝜋2

= 0

1. sin 180°

2. cos 270°

3. sin (−90°)

4. cos(−𝜋)

XVIII. Trig. Identities You should study the following trig identities and memorize them before school starts (we use them a lot)

Find all the solutions to the equations. You should not need a caluclator. (hint one of these has NO solution)

1. sin𝑥 = −12

2. 2 cos𝑥 = √3

3. 4 cos2 𝑥 − 4 cos𝑥 = −1

4. 2 sin2 𝑥 + 3 sin𝑥 + 1 = 0

5. 2 cos2 𝑥 − 1 − cos𝑥 = 0

XIX. Graphing Trig Functions

𝑦 = sin𝑥 and 𝑦 = cos𝑥 have a period of 2𝜋 and an amplitude of 1. Use the parent graphs above to help you sketch a graph of the function below. For 𝑓(𝑥) = 𝐴 sin(𝐵𝑥 + 𝐶) + 𝐾, A = amplitude, 2𝜋

𝐵= period,

𝐶𝐵

= phase shift (positive C/B shift left, negative C/B shift right) and K = vertical shift

Graph two complete periods of the function.

XX. Inverse Trig Functions Inverse Trig Functions can be written in one of two ways:

arcsin(𝑥) sin−1(𝑥) Inverse trig functions are defined only in the quadrants as indicated below due to their restricted domains.

Example 1: Express the value of “y” in radians

𝑦 = arctan−1√3

Solution: Draw a reference triangle This means the reverence angle is 30° or 𝜋

6. So

𝑦 = −𝜋6 so it falls in the

interval from −𝜋2

< 𝑦 < 𝜋2

Thus 𝑦 = −𝜋

6

Example 2: Find the value without a calculator

cos �𝑎𝑟𝑐𝑡𝑎𝑛56�

Solution Draw the reference triangle in the correct quadrant fits. Find the missing side using the Pythagorean

For each of the following, express the value for “y” in radians

1. 𝑦 = arcsin−√32

2. 𝑦 = arccos(−1)

3. 𝑦 = tan−1(−1) For each of the following give the value without a calculator.

4. tan �arccos 23�

Theorem. Find the ratio of the cosine of the reference triangle.

cos𝜃 = �6√61

�

5. sec (sin−1 1213

)

6. sin �arctan 125�

7. sin �sin−1 78�

XXI. Transformations of a Graph Graph the parent function of each set using your calculator. Draw a quick sketch on your paper of each additional equation in the family. Check your sketch with the graphing calculator.

1. Parent Function 𝑦 = 𝑥2 a. 𝑦 = 𝑥2 − 5 b. 𝑦 = 𝑥2 + 3 c. 𝑦 = (𝑥 − 10)2 d. 𝑦 = (𝑥 + 8)2 e. 𝑦 = 4𝑥2 f. 𝑦 = 0.25𝑥2 g. 𝑦 = −𝑥2 h. 𝑦 = −(𝑥 + 3)2 + 6 i. 𝑦 = (𝑥 + 4)2 − 8 j. 𝑦 = −2(𝑥 + 1)2 + 4

2. Parent Function 𝑦 = sin (𝑥) (set mode to radians)

a. 𝑦 = sin (2𝑥) b. 𝑦 = sin(𝑥) − 2 c. 𝑦 = 2sin (𝑥) d. 𝑦 = 2 sin(2𝑥) − 2

3. Parent Function 𝑦 = cos(𝑥)

a. 𝑦 = cos(3𝑥) b. 𝑦 = cos �𝑥

2�

c. 𝑦 = 2cos(𝑥) + 2 d. 𝑦 = −2cos(𝑥) − 1

4. Parent Function 𝑦 = 𝑥3 a. 𝑦 = 𝑥3 + 2 b. 𝑦 = −𝑥3 c. 𝑦 = 𝑥3 − 5 d. 𝑦 = −𝑥3 + 3 e. 𝑦 = (𝑥 − 4)3 f. 𝑦 = (𝑥 − 1)3 − 4

5. Parent Function 𝑦 = √𝑥

a. 𝑦 = √𝑥 − 2 b. 𝑦 = √−𝑥 c. 𝑦 = √6 − 𝑥 d. 𝑦 = −√𝑥 e. 𝑦 = −√−𝑥 f. 𝑦 = √𝑥 + 2 g. 𝑦 = −√4 − 𝑥

6. Parent Function 𝑦 = ln 𝑥 a. 𝑦 = ln(𝑥 + 3) b. 𝑦 = ln 𝑥 + 3 c. 𝑦 = ln(𝑥 − 2) d. 𝑦 = ln−𝑥 e. 𝑦 = −ln𝑥 f. 𝑦 = ln(2𝑥) − 4

7. Parent Function 𝑦 = 𝑒𝑥

a. 𝑦 = 𝑒2𝑥 b. 𝑦 = 𝑒𝑥−2 c. 𝑦 = 𝑒2𝑥 + 3 d. 𝑦 = −𝑒𝑥 e. 𝑦 = 𝑒−𝑥

8. Parent Function 𝑦 = 𝑎𝑥

a. 𝑦 = 5𝑥 b. 𝑦 = 2𝑥 c. 𝑦 = 3−𝑥

d. 𝑦 = 12

𝑥

e. 𝑦 = 4𝑥−3

9. Resize your window to [0,1] × [0,1] Graph all of the following functions in the same window. List the functions from the highest graph to the lowest graph. How do they compare for values of 𝑥 > 1? a. 𝑦 = 𝑥2 b. 𝑦 = 𝑥3 c. 𝑦 = √𝑥

d. 𝑦 = 𝑥23

e. 𝑦 = |𝑥| f. 𝑦 = 𝑥4

10. Given 𝑓(𝑥) = 𝑥4 − 3𝑥3 + 2𝑥2 − 7𝑥 − 11 Use your calculator to find all roots to the nearest 0.0001

11. Given 𝑓(𝑥) = |𝑥 − 3| + |𝑥| − 6 Use your calculator to find all the roots to the nearest 0.001

12. Find the points of intersection. a. 𝑓(𝑥) = 3𝑥 + 2, 𝑔(𝑥) = −4𝑥 − 2 b. 𝑓(𝑥) = 𝑥2 − 5𝑥 + 2, 𝑔(𝑥) = 3 − 2𝑥

(If you found any errors in the packet please let me know so I can correct it. Thanks!)