College Algebra
oNote that x^y computes any number to any power (integer, fraction, decimal).
1)Using the quadratic equation x2 - 3x + 2 = 0, perform the following tasks:
a)Solve by factoring.
Since x 2  3x  2  ( x  2)( x  1)  0 , we get
x-2=0 or x-1=0
So,
x=2 or x=1
b)Solve by completing the square.
x 2  3x  2  ( x  32 ) 2  14 , so,
x 2  3x  2  ( x  32 ) 2  14  0  ( x  32 ) 2 
1
4
So,
x-3/2=1/2 or x-3/2=-1/2
Hence,
x=2 or x=1
c)Solve by using the quadratic formula.
Since x 2  3x  2  0 , by a formula, we have
x

31
2
or x 
31
2
3 32  4*1*2
2*1
So,
x
31
2
i.e.,
x=2 or x=1
2)For the function y = x2 - 6x + 8, perform the following tasks:
a)Put the function in the form y = a(x - h)2 + k.
y  ( x  3) 2  1, so a=1 and k=-1
b)What is the line of symmetry?
The line of symmetry is x=3
c)Graph the function using the equation in part a. Explain why it is not necessary to plot
points to graph when using y = a (x - h) 2 + k.
Show graph here.
Explanation of graphing.
Since we know that the line of symmetry is x=3, we can draw a sketch easily. See
the graph above. You don’t need to have more points to get a sketch
d)In your own words, describe how this graph compares the graph of y = x2?
The graph of y = x2 - 6x + 8 can be obtained from the graph of y = x2. They are
basically the same. You can see the graph of y=x^2 below. Then if we shift this graph to
the right 3 units and then move vertically downwards 1 unit, we can get The graph of
y = x2 - 6x + 8
3)Suppose you throw a baseball straight up at a velocity of 32 feet per second. A function
can be created by expressing distance above the ground, s, as a function of time, t. This
function is s = -16t2 + v0t + s0
·16 represents ½g, the gravitational pull due to gravity (measured in feet per second 2).
·is the initial velocity (how hard do you throw the object, measured in feet per second).
s the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
a)What is the function that describes this problem?
s  16t 2  32t
b)The ball will be how high above the ground after 1 second?
Let t=1, then we have
s  16 *12  32 *1  16( ft)
c)How long will it take to hit the ground?
Let s(t)=0, we have
 16t 2  32t  0
Solving for t, we get
t=2(seconds)
So, it will take 2 seconds to hit the ground
d)What is the maximum height of the ball?
s  16t 2  32t
 16(t  1) 2  16  16
So, the maximum height of the ball is 16 feet.
4)Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a
rectangle is 2 times length plus 2 times width). She wants to maximize the area of her
patio (area of a rectangle is length times width). What should the dimensions of the patio
be, and show how the maximum area of the patio is calculated from the algebraic
equation.
Assume that length is x and width is y. Then we have
2(x+y)=400
So, x+y=200
Now the area is
S=xy=x(200-x)=200x-x^2
It is easy to see that
S   x 2  200 x  ( x  100) 2  10000  10000
When x=100, S achieves its maximum 10000. So, when x=100 and y=200-x=200100=100, the maximum area of the patio is 10000 (ft^2)