VIEWS: 113 PAGES: 39 CATEGORY: Personal Finance POSTED ON: 9/28/2010 Public Domain
Polynomial, Power, and Rational Functions Chapter 2 Linear and Quadratic Functions and Modeling 2.1 Polynomial Polynomials A polynomial, in other words, cannot have x raised to a fractional, zero, or negative power. It also cannot have x in the power. Is the Function a Polynomial? F(x)=-9 + 2x Yes since no powers of x are fractional and x is not a power. F(x)=13 Yes for the same reason as above F(x)=3x No, because x is not the base it is the power. Is the Function a Polynomial? F(x)= x1/2 No, because x is raised to a fractional power. Linear Function A linear function can be written in the form Y=mx+b Finding a Linear Function 1. Find the coordinates. 2. Find the slope or the average rate of change by finding the change in y over change in x. 3. Plug the slope in for m in either y=mx+b or y-y1 = m(x-x1). 4. Plug in one of the points for x and y or x1 and y1, and then solve for b or y. 5. Write out final equation with m and b plugged in. Linear Function, Example Find the equation of a linear function given: F(-3) = 5 and f(6) = -2 Graphing a Linear Function If given two points, then plot each point and connect. Standard Form i Remember that x and y intercepts can help you decide which graph is the graph of the given quadratic function. Match a Parabola to its Function in Vertex Form 1. Find the vertex and look for the graph that has that vertex. 2. Determine if a is positive or negative, so the graph should point up or down. 3. Find the x and y intercepts and check them with the graph. Transformations of a Parabola Remember: -aF(-b(x-h)) + k - Flips over the x-axis A—vertical stretch if a>1, shrink if 0<a<1 - Flips over the y-axis B—horizontal stretch if 0<b<1,shrink if b>1 H—moves left/right K—moves up/down Transformations of a Parabola ¼ x^2 – 1 The ¼ is a and so there is a vertical shrink by ¼. The ¼ is positive, so the graph does not flip over the x –axis. There is no number being added to or subtracted from x, so the graph does not move left or right. The x is not negative so the graph does not flip over the y-axis. The -1 moves the graph down 1. Vertex of a Parabola The point of the parabola that is the max or the min. Find the Vertex and Axis of Symmetry G(x) = -3(x+2)^2 – 1 Vertex is (h,k) or (-2,-1) Axis of symmetry is x=h or x= -2 How to put a Quadratic Function in Vertex Form Use Completing the Square Using Completing the Square to Rewrite a Quadratic Function in Vertex Form Put y or f(x) on the left side of the equation and all other terms on the other side of the equation. Simplify the right side if needed. I will use this example to show what the steps mean. We will rewrite Y=3x2+5x-4 in vertex form, which is Y = a(x-h)2 +k. Using Completing the Square to Rewrite a Quadratic Function in Vertex Form Factor out the coefficient of the x^2 term, a, unless it is one, from all the terms with x on the right side of the equation. In our example, since a is 3 we need to factor a three out from the x^2 and the x terms. Y=3(x2+5/3x)-4 Y = a(x-h)2 +k Using Completing the Square to Rewrite a Quadratic Function in Vertex Form Now, divide the coefficient of the x term by two and then square the result. (5/3)/2 = 5/6 (5/6)2=25/36 Using Completing the Square to Rewrite a Quadratic Function in Vertex Form Add the resulting number to the term in the parenthesis. Note that you are adding ―a‖ times this amount to the function, so you will then need to subtract this same amount so that the function is not changed. In our example we are adding 25/36 to the function 3 times, so we will then subtract 3*25/36. Y=3(x2+5/3x+25/36)-4-3*25/36 Using Completing the Square to Rewrite a Quadratic Function in Vertex Form Factor the quadratic part and simplify the constant part and the function will be in vertex form. Y=3(x + 5/6)2-73/12 Y = a(x-h)2 +k Write an Equation for the Parabola Given the Vertex and a Point Use y=a(x-h)^2 + k to help. 1. Plug the x value of the vertex in as h and the y value in as k. 2. Take the other coordinate and plug in x for x and y for y. 3. Solve this equation for a. 4. Write out the vertex form for the parabola with the a, h, and k plugged in. Modeling • Enter and plot the data as a scatter plot. • Find the regression model that best fits the problem situation. Linear vs. quadratic – Does the scatter look linear or curved? • Place the regression equation into y1, and observe the fit to double check. • Use the regression model to make the predictions called for in the problem. Describing a Linear Correlation • If the scatter points look like they are clustered along a line, then they have a linear correlation. – Positive linear correlation—positive slope – Negative linear correlation—negative slope • r – correlation coefficient—measures the strength and direction of the linear correlation of the data set. Linear Correlation Coefficient Properties • [-1,1] • r>0 there is a positive linear correlation • r<0 there is a positive linear correlation • |r|=1, then there is a strong linear correlation. • r=0, there is a weak or no linear correlation Describing the Strength and Direction of a Linear Correlation 1. State strong or weak based on how linear the points look. The more linear they look, the stronger the linear correlation. 2. Then state positive or negative for direction. Positive slope or negative slope. Quadratic Modeling • Finding maximums and minimums –Calculator • Use max/min function –By hand • Find the vertex and determine if it is a max or a min by determining which way the graph points. Quadratic Modeling Creating Vertical Free-Fall Motion equations. S(t) = -1/2 gt2 +vot+so v(t) = - gt +vo t is time, g=32ft/s2 or 9.8m/s2 is the acceleration due to gravity. Initial height—so Initial velocity--vo Quadratic Modeling Example Fireworks are shot by remote control into the air from a pit that is 10 ft below the earth’s surface. Find an equation that models the height of an aerial bomb t seconds after it is shot upwards with an initial velocity of 80ft/sec. Example Continued We need to find a distance equation. S(t) = -1/2 gt2 +vot+so so = 10 vo = 80 g = 32 ft/sec^2 S(t) = -1/2 (32)t2 +80t+10 S(t) = -16t2 +80t+10 Example Continued What is the maximum height above ground level that the aerial bomb will reach? How many seconds will it take to reach that height? The maximum height will be at the peak of the parabola or the vertex. So use the max function on the calculator and the y value will be the max height and the x will be the time it takes to reach the height. Example Continued What is the maximum height above ground level that the aerial bomb will reach? How many seconds will it take to reach that height? By hand, find –b/2a and this will be the time it takes to reach the height. Plug this value into the function and the result will be the max height.