Transcript Chapter 6 Exploring Quadratic Functions and Inequalities
Chapter 6 Exploring Quadratic Functions and Inequalities
By Jennifer Huss
6-1A Graphing Technology: Quadratic Functions
• Functions with the form y=ax 2 +bx+c are called quadratic functions and their graphs have a parabolic shape • When we solve ax 2 +bx+c=0 we look for values of x that are x-intercepts (because we have y=0) • The x-intercepts are called the
solutions
a quadratic equation or
roots
of • A quadratic equation can have two real solutions, one real solution, or no real solutions
6-1A Graphing Technology: Quadratic Functions (cont.)
• On the calculator find roots using the ROOT menu – Choose a point to the left of the x-intercept and a point to the right of the x-intercept to give a range in which the calculator will find the x-intercept – Do this for each root you see on the graph
6-1A Example
Graph y= -x 2 - 2x + 8 and find its roots.
Vertex: (-1, 9)
Roots: (-4, 0) (2, 0)
Viewing window: Xmin= -10 Xmax=10 Ymin= -10 Ymax= 10
6-1A Problems
1. Find what size viewing window is needed to view y= x 2 + 4x -15. Find the roots.
and 2.3589
: -6.3589 Roots max= 10 Y n= -20 Ymi = 10 Xmax -10 in= : Xm Window
6-1 Solving Quadratic Equations by Graphing
• In a quadratic equation y=ax 2 +bx+c, ax 2 is the quadratic term, bx is the linear term, and c is the constant term • The
axis of symmetry
is a line that divides a parabola into two equal parts that would match exactly if folded over on each other • The
vertex
is where the axis of symmetry meets the parabola • The
roots
or
zeros
(or
solutions
) are found by solving the quadratic equation for y=0 or looking at the graph
6-1 Solving Quadratic Equations by Graphing (cont.)
Graph with definitions shown: Three outcomes for number of roots: Axis of Symmetry Root Root Vertex (2., -5.) Two roots 10 4 2 8 6 -6 -4 2 4 -10 -8 -2 -2 -4 -6 -8 -10 6 8 10 One root: 10 8 6 4 2 -10 -8 -6 -4 -2 -2 -4 -6 -8 -10 2 4 6 8 10 No roots: 10 9 8 7 6 5 4 3 2 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10
6-1 Example
For y= -x 2 -2x + 8 identify each term, graph the equation, find the vertex, and find the solutions of the equation.
-x 2 : quadratic term -2x: linear term 8: constant term Vertex: x=(-b/2a) x= -(-2/2(-1)) x= 2/(-2) x= -1 Solve for y: y= -x 2 -2x + 8 y= -(-1) 2 -(2)(-1) + 8 y= -(1) + 2 + 8 y= 9 -10 Root (-4., 0.) -8 -6 -4 -2 -2 -4 -6 -8 -10 10 Vertex (-1., 9.) 8 6 4 2 Root (2., 0.) 2 4 6 8 10 Vertex is (-1, 9)
6-1 Example (cont.)
Find the roots for the Problem: -x 2 -2x + 8 = 0 (-x + 2)(x + 4) = 0 -x + 2 = 0 x + 4 = 0 -x = -2 x = -4 x = 2 (2, 0) and (-4, 0) are the roots.
6-1 Problems
1. Name the quadratic term, the linear term, and the constant term of y= -x 2 + 4x.
2. Graph y= 4x 2 – 2x + 1 and find its vertex and axis of symmetry.
3. Find the roots of y= x 2 – 8x + 12.
(2,0) and (6,0) 3) ¼ x= (¼, ¾) 2) term onstant no c : linear term 4x c term 2 : quadrati –x 1)
6-2 Solving Quadratic Equations by Factoring
• Factor with the zero product property: if a*b=0 then either a=0 or b=0 or both are equal to 0 • Factoring by guess and check is useful, but you may have to try several combinations before you find the correct one • While doing word problems examine your solutions carefully to make sure it is a reasonable answer
6-2 Example
Solve the equation (2t + 1) 2 – 4(2t + 1) + 3 = 0.
(2t + 1)(2t + 1) – 4(2t + 1) + 3 = 0 4t 2 + 2t + 2t + 1 – 8t – 4 + 3 = 0 4t 2 – 4t = 0 4t (4t – 1) = 0 4t = 0 t – 1 = 0 t = 0 t = 1 The solutions are 0 and 1.
6-2 Problems
1. Solve (5x – 25)(7x + 3) = 0.
2. Solve by factoring: 4x 2 – 13x = 12.
and 4 2) -3/4 and -3/7 1) 5
6-3 Completing the Square
• The way to complete a square for x 2 to take ½ x b and then square it + bx + ? is • So for x 2 + 6x + ? : ½ (6) = 3 3 2 = 9 Therefore, the blank should be 9.
• If the coefficient of x 2 is not 1, you must divide the equation by that coefficient before completing the square • Some roots will be irrational or imaginary numbers
6-3 Example
Find the exact solution of 2x 2 – 6x – 5 = 0.
2x 2 – 6x – 5 = 0 x 2 – 3x – 5/2 = 0 x 2 – 3x + o = 5/2 + o x 2 – 3x + 9/4 = 5/2 + 9/4 (x – 3/2) 2 = 19/4 (x – 3/2) 2 = 19/4 x – 3/2 = + 19/2 or x – 3/2 = - 19/2 Solution: x = 3/2 + 19/2 and x = 3/2 – 19/2
6-3 Problems
1. Find the value c that makes x 2 + 12x + c a perfect square.
2. Solve x 2 – 2x – 15 = 0 by completing the square.
and 5 -3 2) = 36 1) c
6-4 The Quadratic Formula and the Discriminant
• The quadratic formula gives the solutions of ax 2 + bx + c = 0 when it is not easy to factor the quadratic or complete the square • Quadratic formula: x = -b +/- b 2 2a – 4ac • To remember the formula try singing it to the tune of the Notre Dame fight song or “Pop Goes the Weasel”
6-4 The Quadratic Formula and the Discriminant (cont.)
• The b 2 – 4ac term is called the discriminant and it helps to determine how many and what kind of roots you see in the solution
Value of b 2 – 4ac
b2 – 4ac > 0 b2 – 4ac > 0 b2 – 4ac < 0 b2 – 4ac = 0
Is it a perfect square?
yes no not possible not possible
Nature of the Roots
2 real roots, rational 2 real roots, irrational 2 imaginary roots 1 real root
6-4 Example
Find the discriminant of 3x 2 + x – 2 = 0 and tell the nature of its roots. Then solve the equation.
Discriminant = b = 1 2 2 – 4ac – 4(3)(-2) = 1 – (-24) = 1 + 24 = 25 So, there are two real roots and the solutions will be rational.
a = 3 b = 1 c = -2 x = -1 +/- 1 2 – 4(3)(-2) 2(3) x = -1 +/- 25 6 x = -1 +/- 5 6 x = -1 + 5 x = -1 - 5 6 6 x = 2/3 x = -1 The solutions are 2/3 and -1.
6-4 Problems
1. Use the discriminant to tell the nature of the roots of -7x 2 – 8x – 10 = 0.
2. Use the quadratic formula to solve the equation -15x 2 – 8x – 1 = 0.
-1/5 1/3 and 2) nary roots 2 imagi = -216 riminant sc 1) Di
6-5 Sum and Product of Roots
• You can find the quadratic equation from the roots of the equation • If the roots are called S 1 S 1 + S 2 = -b/a and S 1 and S x S 2 2 , then = c/a • This gives us the coefficients of ax 2 + bx + c = 0 • You can also use this method with imaginary roots or to check your solution to a quadratic equation
6-5 Example
Write a quadratic equation from the given roots -4 and -2/3.
-4 + -2/3 = -14/3 -4 x -2/3 = 8/3 a=3 b=14 c=8 3x 2 + 14x + 8 = 0
6-5 Problems
1. Given the roots -1/3 and -1/5, write the quadratic equation.
2. Solve the equation x 2 + 3x – 18 = 0 and check your answers using the sum and product of the roots.
and 3 2) -6 1 = 0 + 2 + 8x 1) 15x
6-6A Graphing Technology: Families of Parabolas
• A parabola has the equation y = a (x – h) 2 + k • The coefficients a, h, and k can be changed to create similar parabolas • Changing “k” moves the parabola up (k > 0) or down (k < 0) • A change in “h” moves the parabola to the right (h > 0) or left (h < 0) • Changing “a” makes a parabola open upwards (a > 0) or downwards (a < 0), and also tells if the parabola is wider ( IaI < 1) or narrower ( IaI > 1)
6-6A Example
Predict the shape of the parabola y = 2 (x+3) 2 + 1 and graph it on a graphing calculator to check your answer.
k = 1 the graph moves up one h = -3 the graph moves three to the left a = 2 the graph is narrower and opens upward
6-6A Problem
1. Predict the shape of y = (x + 2) 2 + 1 and graph the equation on a graphing calculator.
the left o to ed up one and tw 1) Mov
6-6 Analyzing Graphs of Quadratic Functions
• For more information on figuring out the shape of graphs see the notes on 6-6A • The equation y = a (x – h) 2 + k gives the vertex (h, k) and the axis of symmetry is x = h • You can write the equation of a parabola if you know its vertex or if you know three points the parabola passes through
6-6 Examples
1.
Write y = x 2 + 6x – 3 in standard form and then name the vertex, axis of symmetry ,and direction of opening.
y = x 2 + 6x – 3 y + 3 + o = (x 2 y + 3 + 9 = (x 2 + 6x + o + 6x + 9) ) y + 12 = (x + 3) 2 y = (x + 3) 2 – 12 Vertex: (-3, -12) Axis of Symmetry: x = -3 The graph should open upwards.
6-6 Examples (cont.)
2.
Given the points (0, 1) (2, -1) and (1, 3) write the equation of the parabola. Substitute the points into the equation y = ax 2 + bx + c: (0, 1): 1 = a(0) 2 + b (0) + c
1 = c
(2, -1): -1 = a (2) 2 + b (2) + c
-1 = 4a + 2b + c
(1, 3): 3 = a(1) 2 + b (1) + c
3 = a + b + c
Plug in c = 1 for the other two equations: -1 = 4a + 2b + 1
-2 = 4a + 2b
3 = a + b + 1
2 = a + b
6-6 Examples (cont.)
2.
Now solve the system of equations: -2 = 4a + 2b 2 = a + b a = 2 – b -2 = 4 (2 – b) + 2b -2 = 8 – 4b + 2b -2 = 8 – 2b -10 = -2b
b = 5
a = 2 a = 2
a = -3
– b – 5 a = -3 b = 5 c = 1
The equation is y = -3x 2 + 5x + 1.
6-6 Problems
1. Write y = x 2 – 6x + 11 in the form y = a (x – h) 2 + k and find the vertex, axis of symmetry, and direction of opening.
2. Find the equation of the parabola that passes through (0, 0), (2, 6) and (-1, 3). Then graph the function.
Graph of #2 x = 3 y: mmetr of sy is ax (3, 2) ertex: v 2 + 2 ard 2 -x 2x (x –3) opens upw y = 2) y = 1)
6-7 Graphing and Solving Quadratic Equations
• The graph of the parabola serves as a boundary between the area inside the parabola and the area outside the parabola • Graph quadratic inequalities the same way you graph linear inequalities: • Graph the parabola and decide if the boundary line should be solid (≤ or ≥) or dashed ( < or >) • Test one point inside the parabola and one outside the parabola • Shade the region where the inequality was true for the tested points • To solve a quadratic inequality you could graph it or find it through factoring the inequality and testing points
6-7 Examples
1.
Graph the quadratic inequality y > 3x 2 + 12x. Then decide if (2,4) is a solution to the inequality.
Decide where to shade: Test: (0,0) 0 > 3 (0) 2 + 12 (0) 0 > 0 + 0 0 > 0 False Test: (-2, 2) 2 > 3 (-2) 2 + 12 (-2) 2 > 3 (4) – 24 2 > -12 True
Is (2, 4) a solution?
4 > 3 (2) 2 + 12(2) 4 > 12 + 24 4 > 36 You could also look at the graph and see that (2,4) is not in the shaded region.
(2, 4) is not a solution.
6-7 Examples (cont.)
2.
Solve x 2 – 16 < 0.
(x – 4)(x + 4) = 0 x = 4 and x = -4 -4 4 Test in each region so lets choose x = -5, x = 0, and x = 5.
Test: x = -5 (-5) 2 – 16 < 0 25 – 16 < 0 Test: x = 0 (0) 2 – 16 < 0 0 – 16 < 0 Test: x = 5 (5) 2 – 16 < 0 25 – 16 < 0 9 < 0 False -16 < 0 True 9 < 0 False
The solution is -4 < x < 4.
6-7 Problems
1. Graph the quadratic inequality y > x 2 – x + 10 and decide if (0, 12) is a solution of the inequality.
2. Solve x 2 – 10x – 16 < 0.
8 < x 2) 2 < a is ution.
(0, 12) sol 1)
6-8 Integration: Statistics- Standard Deviation
• Standard deviation tells how spread out the values are in a set of data (given symbol σ) • The mean is the average of your data ( symbol x ) • Usually a graphing calculator is used to calculate the standard deviation Standard Deviation = (x 1 – x) 2 + (x 2 – x) 2 + … + (x n – x) 2 n
6-8 Example
Calculate the mean and standard deviation of {3, 5, 6, 7, 9, 11, 22}.
Mean = 3 + 5 + 6 + 7 + 9 + 11 + 22 7 = 9 Standard = Deviation (3 – 9) 2 + (5 – 9) 2 + … + (22 - 9) 2 = 5.8
7
6-8 Problem
1. Calculate the mean and standard deviation of {3, 5, 2, 6, 5, 9}.
2.45
6 or on = Deviati Standard = 5 1) Mean
6-9 Integration: Statistics – Normal Distribution
• A normal distribution curve shows the frequency (how many times something occurs) in a symmetric graph – It is often called a bell-curve because it resembles a bell • Normal Distributions have the following properties: 1. The graph is the highest at the mean 2. The mean, median, and mode are equal 3. Data is symmetrical about the mean
6-9 Integration: Statistics – Normal Distribution
For a Normal Distribution curve: •68% of the values fall within one standard deviation •95% of the values fall within two standard deviations •99% of the values fall within three standard deviations 3 σ 2 σ 1 σ 1 σ 2 σ 3 σ
6-9 Example
• A battery has an average life span of 50 hours, with a standard deviation of 3 hours. The life span of the batteries is normally distributed.
a) What percent of batteries last at least 44 hours?
97.5% of batteries last at least 44 hours.
41 44 47 50 53 56 59 2.5% 95% 2.5% b) If we have 1500 batteries, how many batteries are within one standard deviation of the mean?
68% of batteries are within one standard deviation.
(1500)(0.68) = 1020
1020 batteries are within one standard deviation.