1. For each equation below, identify a,b, and c, the y intercept, the axis of symetry, and tell if it opens up or down. Then graph the function in gcalc.net or other graphing program. The last 2 you will need to get into y = ax^2 + bx + c format
a. y = -3x^2 – 2x + 1
1. a=-3, b=-2, c=1, y intercept = 1, axis of symmetry = -b/2a, or 2/-6, and it opens down since a is negative
b. y = 5x^2 + 3x – 7
1. a=5, b=3, c=-7, y intercept = -7, axis = -3/10, opens up
c. 2y – 5 + 3x^2 = y + x^2 – 3x + 2
solve for y: y = -2x^2 – 3x + 7 now, a = -2, b=-3, c=7, y intercept = 7, axis = 3/-4, opens down
d. 3 + y = 7 – 3x + 2x^2
solve for y, put it ax^2 + bx + c order: y = 2x^2 - 3x + 4 now, a=2, b=-3, c=4, y intercept = 4, axis = 3/4 opens up
2. for each of the above, find the x intercept using the quadratic equation, found in the book. Tell if it actually has x intercepts, and what that means if it does not.
y = -3x^2 – 2x + 1 use quadratic formula: ( -b (+-) (b^2 – 4ac)^(1/2) )/ 2a The easiest way to do this is to find (b^2 – 4ac) first. If this is 0, then there is 1 root at -b/2a, If this is negative, then there are no root, meaning that the graph stays above or below the x axis. If b^2-4ac is positive, the 2 roots are the answer from the rest of the formula: for a above, then, b^2 - 4ac is 4 – 4(-3)(1) or 16.. So, the 2 roots are: -b + 4/2a and -b - 4/2a or 2+4/-6 or -1 and 2 – 4/-6, or 1/3. Plug these 2 values in for x and see that you get y=0.
for problems b: find that b^2 -4ac is 9 – 4(5)(-7) or 9 – (-140) or 149 The square root of 149 is about 12.2. So the roots are: -b + 12.2/2a and -b - 12.2/2a or -3 + 12.2 / 10 and -3 – 12.2 / 10
for problem c. b^2 - 4ac is 9 – 4(-2)(7). or 9 – (-56) or 65. Square root is 8.1. This equation has 2 roots, at 2 + 8.1 / -4 and 2 – 8.1/ -4
for d, b^2 - 4ac is 9 – 4(2)(4) or 9 – 32 or – 23. since the value in the square root is negative, this equation has no roots.
3. Factor: 3(x^3) + 12(x^2) + 9x
First, pull out a common factor of 3 to get 3( x^3 + 4(x^2) + 3x) now pull out a factor of x to get: 3x(x^2 + 4x + 3). Now factor the value inside the ( ) this gives you: (x + 1) and (x + 3) Final answer is 3x(x+1)(x+3). NOTE: I will not ask a factoring problem like the last piece of this on the final quiz. You could us the quadratic equation to get the same answer.
4. Solve: (x + 2)^(1/2) + 2 = 5
isolate the square root to get: (x+2)^(1/2) = 3 square both sides to get: x + 2 = 9. x = 7
5. Find an x that would make this equation true: 3 + 2x = 2x^2 – 3x + 1
Put into ax^2 + bx + c = 0 format. Use the quadratic equation: 0 = 2x^2 – 5x – 2 a=2, b=-5, c=-2 Now find b^2 -4ac. 25 – 4(2)(-2) or 25 – (-16) 41 or 6.4 There are 2 answers: 5 + 6.4 / 4 and 5 – 6.4 / 4
6. Do number 18 for the exercises for 4.2. This is the problem to find the dimensions to enclose the maximum area given you need 3 sides covered (Lake Erie) (A challenge)
check in the textbook
7. Sove: x^(2/3) = 4
take each side of the equation to the 3/2 power. This gives you: x = 4^(3/2) power, or 8