circles standard form: (x – h) 2 + (y – k) 2 = r 2 center: (h, k) radius: r
TRANSCRIPT
Circles
Standard form:
(x – h)2 + (y – k)2 = r2
center: (h, k)
radius: r
Write the equation of the circle.
Writing the Equation of a Circle
(x – 0)2 + (y – 6)2 = 12
x2 + (y – 6)2 = 1
the circle with center (0, 6) and radius r = 1
(x – h)2 + (y – k)2 = r2
Find the zeros of the function by factoring.
Finding Zeros or x-intercepts by Factoring
g(x) = 3x2 + 18x
3x2 + 18x = 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
18x
x(x) – (x) = 3(x)18x Multiply each term by the LCD, x.
x2 – 18 = 3x Simplify. Note that x ≠ 0.
x2 – 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.
24 + 5x = –7x Simplify. Note that x ≠ 0.
24 = –12x Combine like terms.
x = –2 Solve for x.
Solve the equation + = – . 54
6x
74
(4x) + (4x) = – (4x)6x
54
74
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 clubsand a cart at each golf course.
When x = , the y-values are both 102.5. The cost of renting clubs and renting a cart for hours is $102.50 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
x y
60.
60+ =
Write another equation based on the amount of protein:
Protein of beef mix
plus protein of bacon mix
equals
0.18x 0.09y
protein in mixture.
0.15(60)+ =
Solve the system.x + y = 60
0.18x +0.09y = 9
x + y = 60
y = 60 – x
First equation
0.18x + 0.09(60 – x) = 9
0.18x + 5.4 – 0.09x = 9
0.09x = 3.6
x = 40
Solve the first equation for y.
Substitute (60 – x) for y.
Distribute.
Simplify.
Example Continued
x2 – 14x +
Complete the square for the expression. Write the resulting expression as a binomial squared.
Completing the Square
Add.
Factor.
Find .
x2 – 14x + 49
(x – 7)2
Check Find the square of the binomial.
(x – 7)2 = (x – 7)(x – 7)
= x2 – 14x + 49
Add.
Factor.
x2 + 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 + 3x2 = 45
x2 + 6x = 15 Divide both sides by 3.
Simplify.
x2 + 6x + = 15 +
Add to both sides.
x2 + 6x + 9 = 15 + 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) = x2 + 4x + 12
x2 + 4x + 12 = 0
x2 + 4x + = –12 +
x2 + 4x + 4 = –12 + 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 i2 = –1.
–4i + 8i2
–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 i2 = –1.
12 + 24i – 3i – 6i2
12 + 21i – 6(–1)
18 + 21i
Simplify.
Multiply by the conjugate.
Distribute.
Dividing Complex Numbers
Simplify.
Use i2 = –1.
25(x – 2)2 = 9
(x – 2)2 =9
25
x – 2 = 9
25
3
5x – 2
=
EX:
or3
5x – 2 =
3
5–x – 2
=
EXACT: 13/5 and 7/5
Use the quadratic formula to find the real roots of quadratic equations.
x2 + 5x – 1 = 0
a = 1, b = 5, c = –1
x 0.19 or x –5.19
–b b2 – 4ac2a
2(1)
–5 52 – 4(1)(–1)
2–5 29
Solving Quadratics Using the Square Root Property
Subtract 11 from both sides.
4x2 + 11 = 59
Divide both sides by 4 to isolate the square term.
Take the square root of both sides.
Simplify.
x2 = 12
4x2 = 48
Approx: ±3.46
Exact:
Example
Factor: x3 – 2x2 – 9x + 18.
Group terms.(x3 – 2x2) + (–9x + 18)
Factor common monomials from each group.
x2(x – 2) – 9(x – 2)
Factor out the common binomial (x – 2).
(x – 2)(x2 – 9)
Factor the difference of squares.
(x – 2)(x – 3)(x + 3)
Solve the polynomial equation by factoring.
x4 + 25 = 26x2
Set the equation equal to 0.x4 – 26 x2 + 25 = 0
Factor the trinomial in quadratic form.
(x2 – 25)(x2 – 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) = 0x – 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) The set of real numbers between but not including 3 and 5.
-2 -1 0 1 2 3 4 5 6 7 8
3 < 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.
5
22x
–5x –22 Subtract 32.
Divide by –5;reverse inequality sign.
x –22
–5
32 5x 10
Solve a Polynomial Inequality
Solve
x2 – 8x + 15 ≤ 0
(x – 3)(x – 5) ≤ 0
Solve a Polynomial Inequality
f (x) = (x – 5)(x – 3)
Think: (x – 5) and (x – 3) areboth negative when x = –2.
f (x) = (x – 5)(x – 3)
Example 1
Answer: [3, 5]
Solve a Polynomial Inequality