Circles Standard form: (x – h) 2 + (y – k) 2 = r 2 center: (h, k) radius: r

Write the equation of the circle. Writing the Equation of a Circle (x – 0) 2 + (y – 6) 2 = 1 2 x 2 + (y – 6) 2 = 1 the circle with center (0, 6) and radius r = 1 (x – h) 2 + (y – k) 2 = r 2

Find the zeros of the function by factoring. Finding Zeros or x-intercepts by Factoring g(x) = 3x x 3x x = 0 3x(x+6) = 0 3x = 0 or x + 6 = 0 x = 0 or x = –6 Set the function to equal to 0. Factor: The GCF is 3x. Apply the Zero Product Property. Solve each equation.

Key Concept 1

Solve the equation x – = 3. Solving Rational Equations 18 x x(x) – (x) = 3(x) 18 x Multiply each term by the LCD, x. x 2 – 18 = 3x Simplify. Note that x ≠ 0. x 2 – 3x – 18 = 0 Write in standard form. ( x – 6)(x + 3) = 0 Factor. x – 6 = 0 or x + 3 = 0 Apply the Zero Product Property. x = 6 or x = –3 Solve for x.

Multiply each term by the LCD, 4x x = –7x Simplify. Note that x ≠ = –12x Combine like terms. x = –2 Solve for x. Solve the equation + = – x 7 4 (4x) + (4x) = – (4x) 6 x

Solve each equation. The solution x = 2 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution. Divide out common factors. Multiply each term by the LCD, x – 2. Simplify. Note that x ≠ 2. 5x x – 2 3x + 4 x – 2 = 5x x – 2 3x + 4 x – 2 (x – 2) = (x – 2) 5x x – 2 3x + 4 x – 2 (x – 2) = (x – 2) 5x = 3x + 4 x = 2 Solve for x.

City Park Golf Course charges $20 to rent golf clubs plus $55 per hour for golf cart rental. Sea Vista Golf Course charges $35 to rent clubs plus $45 per hour to rent a cart. For what number of hours is the cost of renting clubs and a cart the same for each course? Summer Sports Application

Example Continued Let x represent the number of hours and y represent the total cost in dollars. City Park Golf Course: y = 55x + 20 Sea Vista Golf Course: y = 45x + 35 Because the slopes are different, the system is independent and has exactly one solution. Step 1 Write an equation for the cost of renting clubs and a cart at each golf course.

When x =, the y-values are both The cost of renting clubs and renting a cart for hours is $ at either company. So the cost is the same at each golf course for hours.

A veterinarian needs 60 pounds of dog food that is 15% protein. He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein. How many pounds of each does he need to make the 15% protein mixture? Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.

Example Continued Write one equation based on the amount of dog food: Amount of beef mix plus amount of bacon mix equals xy = Write another equation based on the amount of protein: Protein of beef mix plus protein of bacon mix equals 0.18x0.09y protein in mixture. 0.15(60) + =

Solve the system. x + y = x +0.09y = 9 x + y = 60 y = 60 – x First equation 0.18x (60 – x) = x – 0.09x = x = 3.6 x = 40 Solve the first equation for y. Substitute (60 – x) for y. Distribute. Simplify. Example Continued

x 2 – 14x + Complete the square for the expression. Write the resulting expression as a binomial squared. Completing the Square Add. Factor. Find. x 2 – 14x + 49 (x – 7) 2 Check Find the square of the binomial. (x – 7) 2 = (x – 7)(x – 7) = x 2 – 14x + 49

Add. Factor. x 2 + 9x + Find. Check Find the square of the binomial. Complete the square for the expression. Write the resulting expression as a binomial squared.

Completing the Square – Steps (no decimals - use improper fractions) 1. a must = 1, if not divide everything by a 2. Get the variables on one side and the constant on the other. 3. ADD a blank on both sides 4. Off to the side, take (b/2) 2 (or you can multiply b by ½) 2 5. Put the answer to step 4 in both blanks 6. Factor the LHS using shortcut and simplify the RHS 7. Square root both sides – don’t forget the + and – 8. If you know the square root, set up two equations and solve for x. If not just solve for x.

Solve the equation by completing the square. Solving a Quadratic Equation by Completing the Square 18x + 3x 2 = 45 x 2 + 6x = 15 Divide both sides by 3. Simplify. x 2 + 6x + = 15 + Add to both sides. x 2 + 6x + 9 = Set up to complete the square.

Example Continued Take the square root of both sides. Factor. Simplify. (x + 3) 2 = 24 Exact: Approx: 1.9 and -7.9

Find the zeros of the function by completing the square. Add to both sides. g(x) = x 2 + 4x + 12 x 2 + 4x + 12 = 0 x 2 + 4x + = –12 + x 2 + 4x + 4 = – Rewrite. Set equal to 0. Take square roots. Simplify. Factor. (x + 2) 2 = –8

Express the number in terms of i. Factor out –1. Product Property. Simplify. Express in terms of i. Product Property.

Add or subtract. Write the result in the form a + bi. (4 + 2i) + (–6 – 7i) Add real parts and imaginary parts. (4 – 6) + (2i – 7i) –2 – 5i

Multiply. Write the result in the form a + bi. –2i(2 – 4i) Distribute. Write in a + bi form. Use i 2 = –1. –4i + 8i 2 –4i + 8(–1) –8 – 4i

Multiply. Write the result in the form a + bi. (3 + 6i)(4 – i) Multiply. Write in a + bi form. Use i 2 = – i – 3i – 6i i – 6(–1) i

Simplify. Multiply by the conjugate. Distribute. Dividing Complex Numbers Simplify. Use i 2 = –1.

25(x – 2) 2 = 9 (x – 2) 2 = 9 25 x – 2 =  9 25  3 5  x – 2 = EX: or 3 5 x – 2 = 3 5 – EXACT: 13/5 and 7/5

Use the quadratic formula to find the real roots of quadratic equations. x 2 + 5x – 1 = 0 a = 1, b = 5, c = –1 x  0.19 or x  –5.19 –b  b 2 – 4ac 2a2a 2(1) –5   5 2 – 4(1)(–1) 2 –5   29

Solving Quadratics Using the Square Root Property Subtract 11 from both sides. 4x = 59 Divide both sides by 4 to isolate the square term. Take the square root of both sides. Simplify. x 2 = 12 4x 2 = 48 Approx: ±3.46 Exact:

Example Factor: x 3 – 2x 2 – 9x Group terms. (x 3 – 2x 2 ) + (–9x + 18) Factor common monomials from each group. x 2 (x – 2) – 9(x – 2) Factor out the common binomial (x – 2). (x – 2)(x 2 – 9) Factor the difference of squares. (x – 2)(x – 3)(x + 3)

Solve the polynomial equation by factoring. x = 26x 2 Set the equation equal to 0. x 4 – 26 x = 0 Factor the trinomial in quadratic form. (x 2 – 25)(x 2 – 1) = 0 Factor the difference of two squares. (x – 5)(x + 5)(x – 1)(x + 1) Solve for x. x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0 The roots are 5, –5, 1, and –1. x = 5, x = –5, x = 1 or x = –1

Solve Radical Equations A. Solve. Isolate the radical. x – 5= 0 or x + 1= 0 x = 5 x= –1 You must check your answers

Solve Radical Equations Solve. Isolate a radical. (x – 8)(x – 24) = 0 x – 8 = 0 or x – 24 = 0

Solve the equation. Solving Absolute-Value Equations Rewrite the absolute value as a disjunction. This can be read as “the distance from k to –3 is 10.” Add 3 to both sides of each equation. |–3 + k| = 10 –3 + k = 10 or –3 + k = –10 k = 13 or k = –7

An interval is the set of all numbers between two endpoints, such as 3 and 5. In interval notation the symbols [ and ] are used to include an endpoint in an interval, and the symbols ( and ) are used to exclude an endpoint from an interval. (3, 5)(3, 5) The set of real numbers between but not including 3 and < x < 53 < x < 5

Use interval notation to represent the set of numbers. 7 < x ≤ 12 (7, 12] Interval Notation 7 is not included, but 12 is.

Solve the compound inequality. |2x +7| ≤ 3 Multiply both sides by 3. Subtract 7 from both sides of each inequality. Divide both sides of each inequality by 2. Rewrite the absolute value as a conjunction. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 2x ≤ –4 and 2x ≥ –10 x ≤ –2 and x ≥ –5

Solve linear inequalities in one variable. –5x  –22Subtract 32. Divide by –5; reverse inequality sign. x x  –22 –5 32  5x  10

Solve a Polynomial Inequality Solve x 2 – 8x + 15 ≤ 0 (x – 3)(x – 5) ≤ 0

Solve a Polynomial Inequality f (x) = (x – 5)(x – 3) Think: (x – 5) and (x – 3) are both negative when x = –2. f (x) = (x – 5)(x – 3)

Example 1 Answer: [3, 5] Solve a Polynomial Inequality