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1. Whit’s Wacky GraphsE-Portfolio By: Student 4 7th-Albrecht 

2. Rational Functions Graphing E-Portfolio Word Problem Exponential Function Absolute Value Function Technological Tycoon! Square Root Function Quadratic Function Citations

3. Quadratic Parent Function y=x2 • A parabola is a set of points equally distant from a focus and a directrix. • The general form of a quadratic is y = ax2 + bx + c. When you start graphing your equation is y = a(x – h)2 + k. The a is the same in both equations. You can find your vertex by (h,k) For the parent function of a quadratic equation: • Domain: (- ∞,∞) • Range: ( 0,∞)

4. Graph of Quadratic Parent Function y=x2

5. Reflection of y=x2 y= -x2 Domain: (-∞,∞) Range: ( -∞,0)

6. Vertical Shift y=x2 y=1(x-0)2 +4 Domain: (- ∞, ∞) Range: (4, ∞)

7. Horizontal Shift y=x2 y=(x-4)2 + 0 Domain: (- ∞, ∞) Range: (- ∞, ∞)

8. Graphing Guru of Quadratic Functions • First, you have to make an x and y chart. Plug in an x to find y, then plot the points. • Make sure to do some negative x points so you balance the graph. • I like to find the vertex from the equation first, and then work from there. • Find axis of symmetry by setting the x value of the vertex to x=? • You can tell a lot of things from what a is. For instance, if a is negative then the parabola is upside down. As a varies the curve changes with it.

9. Exponential Functiony=2x • An exponential function is when there is a base a to the x power, that equals f(x). • f (x) = ax • Exponential functions are similar to quadratic equations because they have exponents. However, it is a variable and not a number! For the parent function of an exponential function: • Domain: (- ∞,∞) • Range: ( 0,∞)

10. Graph of Exponential Parent Function y=2x

11. Reflection of y=2x y= 2-x Domain: (- ∞, ∞) Range: (0, ∞)

12. Vertical Shift of y=2x y=2x + 4 Domain: (- ∞, ∞) Range: (4, ∞)

13. Horizontal Shift of y=2x y=2 x y= 2(x-4) + 0 Domain: (- ∞, ∞) Range: (0, ∞)

14. Graphing Guru for Exponential Functions • Well, first, you have to make an x and y chart like every graph. Plug in a number for x to solve for y. • Now, it’s important to remember the rules of exponents. When it is to a negative power remember to put the equation underneath the fraction line. Try to have negative and positive x-values to best represent the graph. • It is probably good to know that exponential functions start off with very small slope and then get bigger as you progress with the graph. They grow very quickly so watch out!

15. Square Root Function y= • A square root function is the opposite of squaring a number. • y= +k • A radical equation is when the unknown variable is stuck inside a radical. The square root symbol is what makes it radical. • For the parent function of radical functions: • Domain: (0,∞) • Range: ( 0,∞)

16. Graph of Square Root Parent Function y=

17. Reflection of Square Root Function y= - Domain: (0, ∞) Range: (- ∞, 0)

18. Horizontal Shift of Square Root Function y= Domain: (-4, ∞) Range:(0, ∞)

19. Vertical Shift of Square Root Function y= + 4 Domain: (0, ∞) Range: (4, ∞)

20. Graphing Guru for Square Root Functions • When dealing with square roots, it is important to consider the domain first. I say this because there cannot be a negative number underneath a square root. To find the domain set everything underneath the square root equal to zero. • Next, start to make your x and y chart. Plug in a value for x and get y! Make sure to pick a widespread of points to get the full effect of the graph!

21. Absolute Value Function y=IxI • The absolute value is the distance from zero. • y=Ix-hI +k • When you take the absolute value of a negative number you get a positive number. For the parent function of absolute value functions: • Domain: (-∞,∞) • Range: ( 0,∞)

22. Graph of Absolute Value Parent Function y=IxI

23. Reflection of Absolute Value Function y=-IxI Domain: (- ∞, ∞) Range: (-∞,0)

24. Horizontal Shift of Absolute Value Function y=Ix+4I + 0 Domain: (- ∞, ∞) Range: (0, ∞)

25. Vertical Shift of Absolute Value Function y=IxI + 4 Domain: (- ∞, ∞) Range: (4, ∞)

26. Graphing Guru for Absolute Value Functions • First, you want to make yourself an x and y table. Put in your equation the value for x to get y. • I like to find the vertex first. You can do this by using the general form of the equation y= aIx-hI+k • However, it is very important to include negative numbers in your chart. You must do this because it may be misleading if you only have positive points. You could mistake it for a linear function when it is not! • Hint--- if you have a number + or – from x inside the absolute value symbol your graph will have a horizontal shift. ( + shifts to the left and – shifts to the left) • Hint--- if you have a number + or – outside of the absolute value symbol your graph will have a vertical shift. ( + shifts up and – shifts down

27. Rational Functiony= 1 x • A rational function is a function of the form f(x)= p(x)/q(x) q(x)≠0 where p(x) and q(x) are polynomial in x. the domain of a rational function consists of the values of x for which the denominator q(x) is not zero. The domain and range for rational parent function: • Domain: (- ∞,0) (0, ∞) • Range: (- ∞, 0) (0, ∞)

28. Graph of Rational Parent Function

29. Reflection of Rational Function y=1/x Domain: (- ∞, 0)(0, ∞) Range: (-∞,0)(0, ∞)

30. Horizontal Shift of Rational Function y=1/(x-4) Domain: (- ∞, 4)(4, ∞) Range: (-∞,0)(0, ∞)

31. Vertical Shift of Rational Function y-4)=1/x Domain: (- ∞, 0)(0, ∞) Range: (-∞,4) (4, ∞)

32. Graphing Guru for Rational Functions • When dealing with rational functions, you must find the asymptotes and domain. Next, construct a table of values, including x-values that are close to the asymptotes on the left and on the right. Plot the points then draw a smooth curve. However, do not connect the two portions of the graph, these are the branches. • An asymptote of a graph is a line to which the graph becomes really close as IxI or IyI increases without bound. • The graph has a vertical asymptote at each real zero of q(x). • The graph has, at most, one horizontal asymptote. • If the degree of p(x) is less than the degree of q(x), then the line y=0 is a horizontal asymptote. • If the degree of p(x) is equal to the degree of q(x), then the line y=a/b is a horizontal asymptote, where a is the leading coefficient in the problem. • If the degree of p(x) is greater than the degree of q(x), then the graph has no horizontal asymptote.

33. Gravel Falling from Conveyor Belt Gravel is falling 4 feet from one conveyor belt to another. Because of the way the conveyor belt is constructed, the gravel is given a slight downward velocity (-1.4 feet per second). How long does it take each piece of gravel to fall from the upper belt to lower belt?

34. Predictions Now that I have my equation h=-16t2 - 1.4t+ 4, I can solve for whatever I like. If it takes two seconds for gravel to fall what will the height be, when the initial height is 4? • First, label what you get from the problem. We know that our t=.46 seconds. Our initial height is s=4, and our height is x. • Now, plug in 2 for t in the equation. • h=-16(2)2 - 1.4(2)+ 4 • Now, solve for h! • You should get -62.8 for the height.

35. Workin’ the Wild Word Problem • First, I wrote down the information given in the problem. I first off knew that I was dealing with a vertical motion problem. So I had to decide if I was going to use h= -16t2 + s, which is height after object is dropped OR h= -16t2 + vt + s, which is height after object is thrown. h=height (feet) t=time in motion (seconds) s=initial height (when t=0) (feet) v=initial velocity (when t=0) (feet per second) • I knew from the information, that this is a quadratic function and I would use the quadratic formula. • Next, I wanted to start labeling my information in the problem. I knew that the time (t) in seconds was my unsolved for variable. I knew my initial velocity (v) was -1.4 feet per second, and I knew my initial height (s) is 4 feet. I know the height (h) is 0 feet. I then made an equation out of this known information. 0= -16t2 - 1.4t + 4. to solve for t, I needed to put it into standard quadratic equation which is ax2 + bx + c =0. So, I had -16t2 -1.4t + 4 = 0. After I had my equation, I labeled my a, b, and c. a= -16 b= -1.4 c= 4. I put those terms in the quadratic formula. –b plus or minus the square root of b squared minus 4ac all over 2a. I solved for t and got approximately .46 feet per second!

36. Graph of Word Problem