Collega Algebre

¶ … Solve the following quadratic equation by factoring:

A) X2 + 6x -16 = 0

(x-2) (x+8)= 0

(x+2) (x-8) = 0

x=-2, x-8

b) solve the quadratic equation 6x2 +3x-18 = 0 using the quadratic formula x= - b +/- ?(b2- 4ac)

+/- ?[32- (4*6*-18c)]

x = 3/2; x= 2

c) Compute the discriminant of the quadratic equation 2x2-3x - 5 = 0 and then write a brief sentence describing the number and type of solutions for the equation.

If x= - b +/- ?(b2- 4ac), then (b2- 4ac) is the discriminant b2- 4ac= -32- (4*2*-5) = 49

There are two solutions for the equation, 1 and 2 1/2, which one gets by plugging the discriminant into the quadratic formula and solving for x.

Use the graph of y=x2+4x-5 to answer the following:

a) Without solving the equation or factoring, determine the solution(s) to the equation, x^2 + 4x - 5 = 0, using only the graph. Answer: x= 0, x = -5. I obtained the answers by looking at the situation where y=0 and finding the value for x in those locations.

b) Does this function have a maximum or a minimum? The graph has a minimum, y= -9, because nothing on the graph goes below that point.

c) What are the coordinates of the vertex in (x, y) form?

(-2,-9)

See below to check:

x= - b +/- ?(b2- 4ac)

2a

y=x2+4x-5

h= -b/2a

h=-4/(2*1)= -2

k= 4ac-b2

4a

k= 4ac-b2 = (4*1*-5) - 42 = -20 -16 = -36/4 = -9

4a

4*1

4

d) What is the equation of the line of symmetry for this parabola? Answer: x= -2

3) The profit function for Wannamaker Trophies is P (x) = -0.4x2 + fx - m, where f represents the design fee for a customer's awards and m represents the monthly office rent. Also, P represents the monthly profit in dollars of the small business where x is the number of awards designed in that month.

a) If is charged $60 for a design fee, and the monthly studio rent is $1,500; write an equation for the profit, P, in terms of x. Type x-squared as x^2

P (x) = -.4x2 + 60x- 1500

b) How much is the profit when 50 award designs are sold in a month? Answer: Show your work here:

P (x) = -.4x2 + 60x- 1500

P (50) = -.4(50)2 + (60* 50)- 1500 = -8,500

c) & d) How many award designs must be sold in order to maximize the profit?

Show your work algebraically. Trial and error is not an appropriate method of solution -- use methods taught in class.

To solve this, I could solve for the vertex, or I could use a graph. I have included the graph below. Solving for the vertex, h is x, which is the number of units sold, and k is P (x), which is th profit.

c) h = -b/2a= -60/(2*-.4)= 75 units

d) k= (4ac-b2)/4a = [(4*-.4*-1500) -- (-602)]/(4*-.4) = 750 maximum profit

c)

d) Answer: Show your work here: