1.  What is a function?  What are the differences between a function and a relationship?  Describe the following types of functions verbally, algebraically and graphically:  Quadratic  Direct variation  Inverse variation  Polynomial  Linear a rule which relates the values of one variable quantity to the values of another variable quantity. It is like a machine. It has an input and an output. A function is a special kind of relation, where no two values of y have the same value of x. A relationship pairs x and y values. Parabola – graph changes direction at the vertex Straight line – graph goes through origin & x = y Varies inversely with x. y or f(x) is inversely proportional to x. Crosses the x-axis up to n times and has n-1 vertices Straight line graph changes at a constant rate See textbook pg. 10

2. HOMEWORK QUESTIONS??????

4.  Transform a given pre-image function so that the result is a graph of the image function that has been dilated by given factors and translated by given amounts  Learn how to transform the equation of a function so that its graph will be dilated and translated by given amounts in the x- and y- directions

5. Dilation translation transformation Inside transformation Outside transformation Image Pre-image Friendly grapher window

6. Solve 1. 2. 3. 4. 5. 6.

7. Each of these two graphs show the unit semicircle and a transformation of it. The left graph shows the semi circle dilated (magnified) by a factor of 5 in the x direction and by a factor of 3 in the y direction. The right graph shows the unit semicircle translated by 4 units in the x-direction and by 2 units in the y-direction. The transformed functions, g and h, in Figure 1-3a are called images of the function f. The original function, f, is called the pre-image.

8. To get the vertical dilation in the left graph Figure 1-3a, multiply each y-coordinate by 3. The figure below shows the image, y = 3f(x).

9.  The horizontal dilation is trickier. Each value must be 5 times what it was in the pre-image to generate the same y-values. Substitute u for the argument of f. y = f(u) Let u represent the original x-values x = 5u The x-values of the dilated image must be 5 times the x-values of the pre- image. 1/5x = u Solve for u y = f(1/5x) Replace u with 1/5x for the argument to obtain the equation of the dilated image.

10.  The figure below shows the graph of the image, y = f(1/5x). Putting the two transformations together gives the equation for g(x) shown in the first figure. g(x) = 3f(1/5x)

11. The equation of the pre-image function in the first figure is f(x) =. Confirm on your grapher that g(x) = 3f(1/5x) is the transformed image function a. By direct substitution into the equation Solution: b. By using the grapher’s built-in variables feature Enter The graph shows a dilation by 5 in the x-direction and by 3 in the y-direction. Use the grid-on feature to make the grid points appear. Use equal scales on the two axes so the graphs have the correct proportions. If the window is friendly in the x-direction (that is, integer values of x are grid points) the the graph will go all the way to x-axis.

12. B. Enter: The graph is the same as the graph of in the third figure. is the function name in this format, not the function value. Note that the transformation in f(1/5x) is applied to the argument of function f, inside the parenthesis. The transformation in 3f(x) is applied outside the parenthesis, to the value of the function. For this reason the transformations are given the names inside transformation and outside transformation, respectively. An inside transformation affects the graph in the horizontal direction, and an outside transformation affects the graph in the vertical direction.

13. You may ask, “Why do you multiply by the y-dilation and divide by the x-dilation?” You can see the reason by substituting y for g(x) and dividing both sides of the equation by 3. You actually divide by both dilation factors, y by the y-dilation and x by the x-dilation. Divide both sides by 3 (or multiply by 1/3

14. The translations in the first figure that transform f(x) to h(x) are shown again in the figure below. To figure out what translation has been done, ask yourself, “To where did the point at the origin move?” As you can see, the center of the semicircle, initially at the origin, has moved to the point (4,2). So there is a horizontal translation of 4 units and a vertical translation of 2 units. To get a vertical translation of 2 units, ad 2 to each y-value:

15. To get a horizontal translation of 4 units, note that what was happening at x = 0 in function f has to be happening at x = 4 in function h. Again substituting u for the argument of f gives: Substitute x – 4 as the argument of f.

16. The equation of the pre-image function in the previous figure is f(x) =. Confirm on your grapher is the transformed image function by a. Direct substitution into the equation b. Using the grapher’s built-in variables feature Solution: Substitute x – 4 for the argument. Add 2 to the expression Enter:

17. The graph for this shows an x-translation of 4 and a y- translation of 2. Again you may ask, “Why do you subtract an x-translation of 4 and a y-translation of 2. The answer again lies in associating the y-translation with the y-variable. You actually subtract both translations: y = 2 + f(x – 4) y – 2 = f(x – 4) The reason for writing the transformed equation with y by itself is to make it easier to calculate the dependent variable, either by pencil and paper or on your grapher.

18. PROPERTY: Dilations and Translations The function g given by or, equivalently, represents a dilation by a factor of a in the y-direction and by a factor of b in the x-direction. The function given by h(x) − c = f (x − d) or, equivalently, h(x) = c + f (x − d) represents a translation by c units in the y-direction and a translation by d units in the x-direction.

19. TEXTBOOK: Pg. 20 #2, 4, 8, 10, 14, 16 JOURNAL QUESTION: Explain in your own words what you learned in this chapter.