美国高中数学1-4 Function

美国高中 数学 数学高考答题卡模板高考数学答题卡模板三年级数学混合运算测试卷数学作业设计案例新人教版八年级上数学教学计划 1-4 Function 4-1: Functions A function is a rule that assigns to every element in a set D (domain) exactly one element in the set R (range). We will treat functions as a set of ordered pairs (x,y) where x is in the domain and y is in the range with y = f(x). Finding the domain of a function from its rule is not difficult. Look at the examples below carefully. 1) Give the domain of each. a) f(x) = 3/(x - 5) The domain consists of all numbers for x that are defined for the function. Since the function doesn't exist at x = 5 ( it makes the denominator 0), the domain is all real numbers but 5. b) f(x) = Since 4 - x is under the radical, 4 - x must be greater than or equal to zero, otherwise the answers will be imaginary. To find the domain, solve the inequality 4 - x > 0. -x > -4 x < 4. Thus, all numbers less than or equal to 4 represent the domain for this function. When trying to find the domain and range from a graph, the domain is found by looking at the graph from left to right. The range is found by looking at the graph from top to bottom. Find the domain and range of the given functions. Look at the graph from left to right. For the x values, they start at -1 and end at 1. So the domain is -1 < x < 1. Look at the graph from top to bottom. The high point is 1 and the low point is 0. Thus the range is 0 < y < 1. The graph above has a domain that is all real numbers. It doesn't stop to the right or the left. The range of the function is y > 0. The low point is zero and it has no high point. To determine if a graph is a function or not, we can use the vertical line test. If no vertical line crosses a graph more than once, then the graph is a function. Study the graphs below and determine if they are functions. .......... .......... The top two are not functions because a vertical line passed through the graphs will cross the circle and parabola twice. The bottom two are functions because vertical lines passed anywhere through these graphs will only cross once. When working with functions, the x variable is the independent variable and f(x) is the dependent variable. The function depends on the values you pick for the x value!! Let's head on to the next section!! Section 4-2: Operations on Functions Operations on functions are similar to operations on numbers. Each operation has a formal definition which produces a new function!! Operations on Functions Each function is defined for all x in the domains of both f and g. 1) Sum of f and g: (f + g)(x) = f(x) + g(x) 2) Difference of f and g: (f - g)(x) = f(x) - g(x) . . 3) Product of f and g: (f g)(x) = f(x) g(x) 4) Quotient of f and g: (f/g)(x) = f(x)/g(x), g(x) not equal to 0 Sample Problems Let f(x) = x + 3 and g(x) = x - 2 1) The sum = (f + g)(x) = f(x) + g(x) = x + 3 + x - 2 = 2x + 1 2) The difference = (f - g)(x) = f(x) - g(x) = (x + 3) - (x - 2) = 5 . .3) The product = (f g)(x) = f(x) g(x) = (x + 3)(x - 2) = 2x + x - 6 4) The quotient = (f/g)(x) = f(x)/g(x) = (x + 3)/(x - 2), x does not = 2 Another way of combining functions is the composite function. The composite function is denoted by: and is = f(g(x)). x must be in the domain of g(x) and g(x) must be in the domain of f(x). Example 1 Find f(g(x)) and g(f(x)) for the functions f(x) = x + 2 2and g(x) = x 22 f(g(x)) = f(x) = x + 2. The function g(x) squares any number. The function f(x) simply adds two to any number. G(x) is applied first (it squares x) and then f(x) is applied (it adds two). The domain is all real numbers. 2 g(f(x)) = g(x + 2) = (x + 2) This time, f(x) is applied first and then g(x). Take any number and add two first then square the result. The domain is all real numbers. Notice that the answers are not the same illustrating that the composite function is not commutative. This means the order in which the problem is written is important. Example two Find f(g(x)) and g(f(x)) for the functions f(x) = 1/x and g(x) = 2x. f(g(x)) = f(2x) = 1/(2x). The domain is all real numbers but zero. g(f(x)) = g(1/x) = 2/x. The domain is again all real numbers but zero. Example three Find f(g(x)) and g(f(x)) for the functions f(x) = x + 3 and g(x) = 2/x f(g(x)) = f(2/x) = 2/x + 3. The domain is all real numbers but 0. g(f(x)) = g(x + 3) = 2/(x + 3). The domain is all real numbers but -3. Sample Problems 221) On the same axes graph f(x) = x, g(x) = x + 3, and 2h(x) = x - 2 The effect of adding three, simply moved the graph up 3 units while the effect of subtracting two simply moved the graph down two units. 22) Let f(x) = 4x, g(x) = 3x + 2 and h(x) = x Find f(g(h(3))) = f(g(9)) = f(29) = 116 Find h(g(f(-1))) = h(g(-4)) = h(-10) = 100 Find h(h(h(2))) = h(h(4)) = h(16) = 256 Find h(g(f(x))) = h(g(4x)) = h(12x + 2) = (12x + 22) Let's head over to the next section! I think I need to head back! Section 4-3: Reflections and Symmetry Try the quiz at the bottom of the page! go to quiz Reflections A reflection is like a mirror image. The line of reflection acts as the mirror and is halfway between the point and its image. If the point is its own reflection, then it is a point on the line of reflection. Some of the important relections are listed below with examples. 1) y = -f(x) (This is the reflection about the x-axis of the graph y = f(x).) That is for every point (x, y) there is a point (x, -y). 22 Look at the example graphs below of y = x and y = -x. Notice the green graph is simply the same as the blue graph folded down across the x-axis. The x-axis is the line of reflection or the mirror. 2) y = |f(x)| means that the entire graph will be above the x-axis. Why? (The absolute value is always positive, that's why!!) To graph the absolute value graph, graph the function y = f(x). Anything above the x-axis, stays above it, anything below the x-axis is reflected above the x-axis and anything on the x-axis, stays on the x-axis. Study the following graphs that illustrate: .................... 2 y = x - 4 y = 2|x - 4| The absolute value of the left graph is found by taking all points above the x-axis and leaving them above the x-axis. These are the purple points. The points below the x-axis are reflected up above the x-axis. These are the red points. Of course, the points on the x-axis stay on the x-axis. 3) y = f(-x) (This is reflection about the y-axis of the graph y = f(x)) For every point on the right of the y-axis, there is a point equidistant to the left of the y-axis. That is for every point (x, y), there is a point (-x, y). Study the graphs below: Notice that every point on the left of the y-axis has an image on the right of the y-axis. Look at the equations. The only thing that makes them different is the negative sign in front of only the x variable. That is what makes them reflections about the y-axis. 4) Reflections about the line y = x is accomplished by interchanging the x and the y-values. Thus for y = f(x) the reflection about the line y = x is accomplished by x = f(y). Thus the reflection about the line y = x for y = 22 is the equation x = y. These are graphed on the x following graph: The blue line is where you would fold the paper for these two graphs to match. Notice that the green parabola is opening up but when you fold it across the blue line, it opens to the right. Symmetry A line is called an axis of symmetry of a graph if you can pair the points such that the line is perpendicular of the segment joining the pair of points. Study the graph below to identify the axis of symmetry: Ways to test different symmetry Types of symmetry Symmetry to the x-axis Test for symmetry (x, -y) is on the graph In the equation, leave x alone and replace y with -y. If you get when (x, y) is. the same equation, it is symmetric to the x-axis. Symmetry to the y-axis Test for symmetry. (-x, y) is on the graph In the equation, replace x with -x. If the equation is the same, it when (x, y) is. is symmetric to tye y-axis. Symmetry to the line y = Test for symmetry. x. Interchange x and y. If the equation is the same then it is (y, x) is on the graph symmetric to the line y = x. when (x, y) is. Symmetry to the origin. Test for symmetry. (-x, -y) is on the graph Replace x with -x and y with -y. If the equation is the same, then when (x, y) is. it is symmetric to the origin. Examples 221) Find the symmetry for the equation x + y = 9. Symmetry to Symmetry to Symmetry to y = Symmetry to x-axis y-axis x origin 22222222x + (-y) = 9 (-x) + y = 9 y + x = 9 (-x) + (-y) = 9 22222222 x + y = 9 x + y = 9 x + y = 9 x + y= 9 42) Find the symmetry for the equation x = y - 3. Symmetry to Symmetry to Symmetry to y = Symmetry to x-axis y-axis x origin 4 44x= (-y) - 3 (-x) = y - 3 (-x) = (-y) - 3 4y = x - 3 444 x = -y - 3 x = y - 3 x= -y - 3 Graphing Problems 1) Graph (blue) Find and graph the equation of : a) Reflection about x-axis (green) b) Reflection about y-axis (purple) c) Reflection about y = x (red) Solutions: 2) Use symmetry to graph |x| + |y| = 1 Symmetry to Symmetry to Symmetry to y = Symmetry to x-axis y-axis x origin. |x| + |-y| = 1 |-x| + |y| = 1 |y| + |x| = 1 |-x| + |-y| = 1 |x| + |y| = 1|x| + |y| = 1|x| + |y| = 1|x| + |y| = 1 yes yes yes yes Because we have symmetry to all four, we have a reflection in quadrant II, III and IV. Thus, all you need to do is make the graph in Quad I and reflect into the other three quadrants. The equation in quad I is x + y = 1 Now reflect into the other three quadrants: Sail towards the next section: Sail back to the previous section: <> Current quizaroo # 4a 1) Give the domain for the function: f(x) = (x - 3)/(x - 1) a) All real numbers but 3 and 1 b) All real numbers c) All real numbers but 3 d) All real numbers but 1 e) All real numbers between 1 and 3 2) Give the range for the function: f(x) = | x | + 1 a) y > 0 b) y > 1 c) All real numbers d) x > 0 e) x > 1 23) Given f(x) = x, g(x) = 2x and h(x) = x + 3, find f(g(h(2))) = a) 100 b) 11 c) 49 d) 50 e) 19 24) Tell the symmetry for y + xy = 10. Use a) x-axis, b) y-axis, c) line y = x, d) origin a) a only b) b only c) c only d) d only e) not symmetric to the above four 5) Classify the function f(x) = | x | a) odd function b) even function c) both even and odd d) neither even or odd e) symmetric to origin click here for answers!! Section 4-4:Periodic Functions - Stretching and Translating A function f is periodic if there is a positive number p such that: f(x + p) = f(x) for all x in the domain of f. The definition means that the y values will repeat over some p value called the fundamental period of the function. Look at the graph: The graph starts at 0, goes up to 2, back down to 0, down to -2 and back to 0. At this point the graph starts repeating. Look at the x value to find the period length. The period length p = 4, because it takes 4 units for this graph to repeat the y values. You can tell where the graph will be at larger values by knowing the repeat. f(0) = 0, f(1) = 2, f(2) = 0, f(3) = -2 and f(4) = 0. What is f(21)? Divide 21 by 4 and use the remainder. The remainder is 1. Thus f(21) = f(1) = 2 What is f(82)? Divide 82 by 4. The remainder is 2. Thus f(82) = f(2) = 0. If a periodic graph has a maximum value M and a minimum value m, then the amplitude A of the function is: A = (M - m)/2 The amplitude of the above graph is: A = (2 -(-2))/2 = 4/2 = 2 Stretching a graph vertically The graph of y = cf(x) where c is a positive number not equal to 1, is obtained by vertically stretching or shrinking the graph of y = f(x). Let f(x) be the following graph: Now compare these two graphs to the green one above. The purple graph doubled the green graph. Notice, all that changed was the high and low points of the graph. In other words, stretched vertically. Look at the red graph. It is the green graph multiplied by 1/2. All that has changed is the high and low points. In other words, shrunk vertically. The period length has not changed. In all three graphs, the period length is 3. Horizontally stretching or shrinking The graph of y = f(cx) where c is positive and not equal to one is obtained by horizontally stretching or shrinking the graph of y = f(x). If c > 1 it is a horizontal shrink. If 0 < c < 1, it is a horizontal stretch. Here is y = f(x): Watch the effect of multiplying the x value by 2. Notice the amplitude didn't change. The graph high and low points are the same. But look at the purple graph. Notice it has gone through two complete cycles by the time the green graph has gone through one cycle. It is like compressing a spring. Now watch the effect of multiplying by 1/2. The effect this time is to stretch the graph. Look at the red graph. At x = 3, the red graph is only half way through the cycle. It takes 6 units for the red graph to repeat instead of three for the green graph. It is like pulling out on a spring. Summary of above If a periodic function f has a period p and amplitude A, then: y = cf(x) has period p and amplitude cA y = f(cx) has period p/c and amplitude A Translating Graphs A translation is simply moving the exact same graph to another location. The size and shape does not change from the original graph, only the placement of the graph changes. Your knowledge of basic graphs is very helpful when doing translations. Here is how to translate: y - k = f(x - h) is obtained by shifting the graph of y = f(x), k units up/down and h units right/left. 2You already know what y = x is. How does y - 2 = (x - 21) compare? Notice the green graph is the same size and shape of the blue graph. It is shifted one unit right and two units up. Now graph y + 1 = | x + 2|. This depends on you knowing that the absolute value graph is a v-shaped graph. So this is a translation of y = |x| The green graph is the graph of the function we want. It is a translation of the blue graph moved one unit down and 2 units left. Notice in this problem and the last problem what causes the graph to be shifted right vs. left and up vs. down. Combining reflections and translations When combining reflections and translations, remember to reflect first then translate. Failure to work the problem in this order may result in the wrong answer. Graph the function y - 1 = -|x + 1| The basic graph is y = |x|, a v-shaped graph. The negative sign in front makes this a reflection about the x-axis. Do this first. Then translate the result by moving the graph up one and one to the left. Our answer is in purple. Remember to make a copy of the chart on page 142 in your notebook. You must know that chart!! It helps a great deal in future chapters!! Let's shift into the next section: Let's reflect back on the previous section: Section 4-4:Periodic Functions - Stretching and Translating A function f is periodic if there is a positive number p such that: f(x + p) = f(x) for all x in the domain of f. The definition means that the y values will repeat over some p value called the fundamental period of the function. Look at the graph: The graph starts at 0, goes up to 2, back down to 0, down to -2 and back to 0. At this point the graph starts repeating. Look at the x value to find the period length. The period length p = 4, because it takes 4 units for this graph to repeat the y values. You can tell where the graph will be at larger values by knowing the repeat. f(0) = 0, f(1) = 2, f(2) = 0, f(3) = -2 and f(4) = 0. What is f(21)? Divide 21 by 4 and use the remainder. The remainder is 1. Thus f(21) = f(1) = 2 What is f(82)? Divide 82 by 4. The remainder is 2. Thus f(82) = f(2) = 0. If a periodic graph has a maximum value M and a minimum value m, then the amplitude A of the function is: A = (M - m)/2 The amplitude of the above graph is: A = (2 -(-2))/2 = 4/2 = 2 Stretching a graph vertically The graph of y = cf(x) where c is a positive number not equal to 1, is obtained by vertically stretching or shrinking the graph of y = f(x). Let f(x) be the following graph: Now compare these two graphs to the green one above. The purple graph doubled the green graph. Notice, all that changed was the high and low points of the graph. In other words, stretched vertically. Look at the red graph. It is the green graph multiplied by 1/2. All that has changed is the high and low points. In other words, shrunk vertically. The period length has not changed. In all three graphs, the period length is 3. Horizontally stretching or shrinking The graph of y = f(cx) where c is positive and not equal to one is obtained by horizontally stretching or shrinking the graph of y = f(x). If c > 1 it is a horizontal shrink. If 0 < c < 1, it is a horizontal stretch. Here is y = f(x): Watch the effect of multiplying the x value by 2. Notice the amplitude didn't change. The graph high and low points are the same. But look at the purple graph. Notice it has gone through two complete cycles by the time the green graph has gone through one cycle. It is like compressing a spring. Now watch the effect of multiplying by 1/2. The effect this time is to stretch the graph. Look at the red graph. At x = 3, the red graph is only half way through the cycle. It takes 6 units for the red graph to repeat instead of three for the green graph. It is like pulling out on a spring. Summary of above If a periodic function f has a period p and amplitude A, then: y = cf(x) has period p and amplitude cA y = f(cx) has period p/c and amplitude A Translating Graphs A translation is simply moving the exact same graph to another location. The size and shape does not change from the original graph, only the placement of the graph changes. Your knowledge of basic graphs is very helpful when doing translations. Here is how to translate: y - k = f(x - h) is obtained by shifting the graph of y = f(x), k units up/down and h units right/left. 2You already know what y = x is. How does y - 2 = (x - 21) compare? Notice the green graph is the same size and shape of the blue graph. It is shifted one unit right and two units up. Now graph y + 1 = | x + 2|. This depends on you knowing that the absolute value graph is a v-shaped graph. So this is a translation of y = |x| The green graph is the graph of the function we want. It is a translation of the blue graph moved one unit down and 2 units left. Notice in this problem and the last problem what causes the graph to be shifted right vs. left and up vs. down. Combining reflections and translations When combining reflections and translations, remember to reflect first then translate. Failure to work the problem in this order may result in the wrong answer. Graph the function y - 1 = -|x + 1| The basic graph is y = |x|, a v-shaped graph. The negative sign in front makes this a reflection about the x-axis. Do this first. Then translate the result by moving the graph up one and one to the left. Our answer is in purple. Remember to make a copy of the chart on page 142 in your notebook. You must know that chart!! It helps a great deal in future chapters!! Let's shift into the next section: Let's reflect back on the previous section: Section 4-5: Inverse Functions An inverse is the operation that takes you back to where you started. The inverse of multiplication is division, adding and subtracting, square and square root, etc. For functions, there are two conditions for a function to be the inverse function: 1) g(f(x)) = x for all x in the domain of f 2) f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x. The notation used to indicate an inverse function is: f -1(x) pronounced "f inverse". This notation does not mean 1/f(x). Example 1) If f(x) = 3x - 1 and g(x) = (x + 1)/3, show that f and g are inverses to each other. To show that they are inverses, we must prove both of the above parts. g(f(x)) = g(3x - 1) = (3x - 1 + 1)/3 = 3x/3 = x f(g(x)) = f((x + 1)/3) = 3[(x + 1)/3] - 1 = x + 1 - 1 = x Since both parts work, they are indeed inverses of each other. To find a rule for the inverse function Find the inverse function for y = 5x + 2 To find the inverse, interchange x and y. x = 5y + 2 Now isolate for y!! x - 2 = 5y (x - 2)/5 = y We now have the inverse!! Notice, that this inverse make sense. The original problem had adding by two and the inverse is subtracting two. The original function had multiplying by five and the inverse has division by five. Graphs of inverse functions We have to make sure that the inverse is indeed a function. Not all functions will have inverses that are also functions. In order for a function to have an inverse, it must pass the horizontal line test!! Horizontal line test If the graph of a function y = f(x) is such that no horizontal line intersects the graph in more than one point, then f has an inverse function. This will make sense when we discover how to graph the inverse function. To graph the inverse function, it is simply the reflection about the line y = x. Makes sense, because in order to get the graph, we interchange x and y. Recall from previous sections what the reflection about the line y = x looks like. Any two points on the same horizontal line when reflected will be on the same vertical line. Can't have this because it wouldn't be a function. That's why the horizontal line test works. Example -1-11) Find the equation of f and graph f, f , and y = x for f(x) = 2x - 5. First, f(x) is a line and it passes the horizontal line test. Find the inverse: y = 2x - 5 x = 2y - 5 x + 5 = 2y (x + 5)/2 = y 22) Let f(x) = 9 - x for x > 0 -1 Find the equation for f (x) -1 Sketch the graph of f, f , and y = x. Notice that the equation is half of a parabola. Only the side to the right of zero. If we tried to use the entire parabola, it wouldn't pass the horizontal line test. 2 To find the equation: y = 9 - x 2 x = 9 - y 2 x - 9 = -y 2 9 - x = y , x < 9 On to the last section: Back up and regroup: Section 4-6: Functions of Two Variables Try the quiz at the bottom of the page! go to quiz Suppose a company makes two products. Let x represent the number of the first product made and y represent the number of the second product made. If the first product makes $2 each and the second $3 each, the profit function would be: P(x, y) = 2x + 3y Notice that profit is a function of two variables. There are a few ways we could graph this function. One way is to graph in a three dimensional plane. We will talk about this later in the year. Another way to graph this function is to hold one of the variables constant. For example, we could hold y constant like: y = 2, or y = 3 or y = 4, etc. The equations would look like: P(x, 2) = 2x + 6 P(x, 3) = 2x + 9 P(x, 4) = 2x + 12 These graphs all have the same slope but different y-intercepts. These make a family of curves. They will be graphed in the xP plane. Another way we can graph this is by holding the profit constant. For example, we could let P = 6 or P = 12 or P = 18. Then the equations would be: 6 = 2x + 3y 12 = 2x + 3y 18 = 2x + 3y Another equation we could use is: d(r, t) = rt Suppose we hold t constant at t = 1, t = 2 and t = 3. Make a graph of these three equations: d(r, 1) = r d(r, 2) = 2r d(r, 3) = 3r Now graph the same function holding the distance constant at d = 10, d = 20, and d = 30. The equations are 10 = rt 20 = rt 30 = rt I'm ready for that practice test now! I'm almost ready. I think I will go back and review! Current quizaroo # 4b 1) y + 3 = -f(x - 1) Describe the translations, symmetry and stretching/shrinking for this graph: a) Reflected about the y-axis, moved three units up and one unit to the left b) Reflected about the y-axis, moved three units down and one unit to the right c) Reflected about the x-axis, moved three units up and one unit to the left d) Reflected about the x-axis, moved three units down and one unit to the right e) Moved three units down and one unit to the right 2) Write the equation of a function that is shifted 2 units up and 4 units to the left, that is reflected about the x-axis with a stretch vertical factor of 2. a) y - 2 = -2f(x + 4) b) y + 2 = -2f(x - 4) c) y - 2 = 2f(x + 4) d) y + 2 = 2f(x - 4) e) y +4 = -2f(x - 2) 3) Find the inverse function to the equation: y = -3x + 4 a) y = 3x - 4 b) y = (4 - x)/3 c) y = (x - 4)/3 d) y = (x + 3)/4 e) y = (3 - x)/4 24) If g(x) = x - 2, x < 0, find the inverse function. ____ a) y = -\/ x + 2, x > -2 ____ b) y = \/ x + 2, x > -2 c) y = x + 2, x > -2 d) y = -(x + 2), x > -2 e) doesn't have an inverse function, because it doesn't pass the horizontal line test 5) Name an ordered pair that is not in the domain of f(a, b) = (a - b)/(a + b) a) (5, 5) b) (x, x) c) (-d, d) d) (-4, -4) e) (3/2, 2/3) click here for answers!! Sample test!! 1) Give the domain, range and zeros of the following graph: 2) Give the domain, range and zeros of the function: f(x) = 5x/(3x - 2) 3) Sketch the graph and use the graph to find the 32range and zeros of the function: f(x) = x + 4x - x - 4 24) Let f(x) = 4x, g(x) = x, h(x) = x + 5 and j(x) = 2x. Find each. a) (f + h)(x) b) (j - f)(x) c) f(g(h(x))) d) g(h(j(x))) e) (f/g)(x) f) g(f(h(2))) 225) Sketch the graphs of y = x -4, y = 4 - x and y = |4 - 2x| on the same axes. 6) Test each equation for all 4 symmetries. 2 a) x - xy = 3 22 b) x - y = 1 c) y = -|x| 27) Use symmetry to sketch the graph of xy = 2 8) Sketch each of the following by using stretching/shrinking and/or translation. a) y - 5 = 2|x - 1| 2 b) y + 3 = -(x + 1) 9) Prove that f(x) = (x+1)/2 and g(x) = 2x - 1 are inverse functions. -110) Sketch the graph of f and f . Then find a rule for the inverse. 2f(x) = x - 2, x < 0 11) a) Let P(x, y) = 2x + 2y. Let y be constant at y = 2, y = 3, y = 4. Graph these functions. b) Now let P be constant at P = 10, P = 12 and P = 14. Graph these functions. 12) The area of a triangle is a function of the base b and height h. a) Express A as a function of b and h. b) Find A(2,3) and A(5,6) c) Draw the curve of constant area A(b, h) = 5 in a bh-plane. 13) let f(x) be the following graph: a) What is the fundamental period for the above graph? b) What is f(1)? f(2)? f(3)? f(4)? c) What is f(99)? f(13)? f(-2)? f(-11)? d) Graph 2f(x) e) Graph f(2x) f) Graph -f(x) g) Graph f(-x) This page hosted by Get your own Free Home Page Hope you did well! Hit me to see the answers!! I didn't do well. Send me back!! Answers To Sample Test 1) Domain -- x < 4 Range -- y > 0 Zero at ( 4, 0) 2) Domain -- all real numbers but 2/3 Range -- all real numbers but 5/3 Zero at ( 0, 0) 3) Range is all real numbers zeros at (-4, 0), (-1, 0), and (1, 0) 4) a) 5x + 5 b) -2x 2 c) 4(x + 5) 2 d) (2x + 5) e) 4/x, x can't equal 0 f) g[f(7)] = g(28) = 784 5) a) b) c) 6) x y y = x origin a) no no no yes b) yes yes no yes c) no yes no no 7) It is symmetric to the x-axis 8) a) is a vertical stretch and translation of 1 right and 5 up b) is a reflection about the x-axis and a translation of 1 left and 3 down 9) To prove this you must show both ways!! f[g(x)] = f(2x-1) = (2x - 1 + 1)/2 = 2x/2 = x g[f(x)] = g[(x + 1)/2] = [2(x + 1)/2] - 1 = x + 1 - 1 = x 10) To find the rule interchange x and y and solve for y. 2 x = y - 2 2 x + 2 = y _____ y = - \/ x + 2 , x > -2 11) a) b) 12) a) A(b, h) = bh/2 b) A(2, 3) = (2)(3)/2 = 3 and A(5, 6) = (5)(6)/2 = 15 c) 5 = bh/2 ==> 10 = bh ==> h = 10/b 13) a) 4 b) 1, 0, -1, 0 c) -1, 1, 0, 1 d) Changes amplitude but not period length. e) Changes period length to 2 (horizontal shrink) but does not change amplitude f) Reflection about the x-axis g) Reflection about the y-axis This page hosted by Get your own Free Home Page Hope you did well on this opportunity. If not go back and study the previous sections until you have mastered the material. Especially study the reflections, stretches and shrinks, and of course, the translations. Good luck!!