1. Since you already have an x value of 1 with the first point, you cannot have a different y value for x with the third point. Therefore, the blank MUST be a 3, or the function would have different values when x=1, and therefore would not be a function.
2. The vertical line test can be used to see if a graph represents a function. If you draw a vertical line through a graph and it intersects 2 points on the graph, the graph cannot be a function. This is because the graph would have 2 different values for the same x, violating the rules for a function
3. Put this into y= mx + b format by solving for y You get: 2y = -3x + 8. solve for y to get y = -3/2(x) + 4 Slope is -3/2
4. y = 3 - | x+1| The smallest value for the absolute value is when x = -1.. This will get you a 0. All other times it will be bigger. When x=-1, then, you will have the ’point’ of the V. Plug in and see that at this point, y will be 3. Now check a point on either side of -1.. at 0, y = 2, so you have (0,2), and when x=-2 you have (-2,2). Notice the V is up side down, since you are subtracting ever bigger numbers.
5. y = 2x + b. Solve for b by plugging in your point for x and y. This gives youL 5 = 2(3) + b or 5 = 6 + b. Subtract 6 from both sides to get b=-1… The equation is y = 2x -1 (or y = 2x + -1)
6. Since they have the same slope, and have different y intercepts, they must be parallel.
7. (-3)(4)/(-6) – (2)(-3) + (-1)(3) – (-1)(1) This problems is checking for 2 things.. You need to multiply and divide first, and you need to know the rules for negative numbers;
-12/(-6) – (-6) + (-3) – (-1) or 2 – (-6) + (-3) – (-1) or 2 + 6 – 3 + 1 or 6
8. A practical range is the realistic y values that are capable for a function.
9. Find (2f – g) (x) f(x) = 3(x ^2) – 2 and g(x) = 4x – 1
2f = 2(3(x^2)-2 = 6x^2 – 4
2f-g = 6x^2 – 4 – (4x – 1) or 6x^2 – 4x – 3
10. 3g(x) = 3(4x-1) or 12x – 3 So.. find (3x^2 – 2)(12x-3)
= 36x^3 – 9x^2 – 24x + 6
11. 3(4x-1)^2 – 2 or 3(4x-1)(4x-1) – 2 or 3(16x^2 – 8x + 1) -2
or 48x^2 – 24x + 3 – 2 or 48x^2 – 24x + 1
12. Find the inverse of y = 4x-1 swap x and y and solve for y: x = 4y-1 4y = x+1 y = (x+1)/4
13. 3x + 2 – 3y/2 = 5 – x + y Get the y’s on the left and x’s and numbers on the right: -3y/2 – y = -4x+ 3 Multiply both sides by 2 to get rid of the bottom 2 in the fraction: 2(-3y/2 – y) = 2(-4x + 3) or -3y – 2y = -8x + 6 or -5y = -8x + 6 or y = (-8x+6)/-5 or, if you like: y = 8x/5 – 6/5
14. (x^3 – 3x)(2x^2 – 4x + 1) or 2x^5 – 4x^4 + x^3 – (6x^3 – 12x^2+ 3x) or 2x^5 – 4x^4 + x^3 – 6x^3 + 12x^2 – 3x or 2x^5- 4x^4 – 5x^3 + 12x^2 – 3x
15. x^0 =1, y^0 = 1 answer: 2
16. x to the 5th power is on the inside of the root sign. 2 is outside the root sign
17. 2(3)(1)(12) / (6)(6)(2)(52)
18. 5 < 3 + 2x/3 < 11 or 2 < 2x/3 < 8 or 6 < 2x < 24 or 3 < x < 12
19. x^17
20. x^(-3/4)/(x^(2/3)) subtract powers, so x^(-3/4 – 2/3) or x^(-9/12 – 8/12) or x^(-17/12)
21. .00000000481 or 4.81(10^-9)
22. The square root of 4^3 / square root of 4, or the square root of 64 / 2 or 8/2 or 4
23. 58,698 < x < 58.702
24. A piecewise function is one that has different functions of x for different domains of x, with no overlaps
25. Simple interest: A = Pr^t or A = 5000(1.14)^5
26. You want to find the time it takes to get to 4 acres. A = 2(1.18)^t
Plug in different values for t until A = 4 (a little over 4 hours)
27. A = 10000(1 + .05/12)^12(10) is the yield if compounded monthly. You want the value in the equation A = 10000(r)^10
This is the simple interest rate, if just compounded at the end of the year. So, if you compare the 2 equations, notice that r, in the second equation, the effective yield, is (1+.05/12)^12 (all the stuff between the 10000 and the last (10) in the first equation) Find this value to be: 1.051162 about. So the the effective yield is about 5.1162 percent.
28. 5000e^(.1)(15) 0r about $22,429
29. 4, at most
30. As x gets more and more negative, x^3 will get more and more negative, so the graph just keeps going down as you go farther to the left.