Example: Solve the equation.
Solution:
Example: Solve the equation.
Solution:
Example:
A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by the quadratic function . How long does it take the baseball to reach its maximum height? What is the maximum height obtained by the baseball?
Solution:
is defined to be a quadratic function. The graph of a quadratic function is a parabola. This quadratic function has a leading coefficient of 4.9 which means that the parabola is opening down.
Find the vertex to find the maximum value.
The ball reaches the maximum height 1.5 seconds after the ball was hit.
The maximum height can be found by substituting 1.5 seconds for time in the height function.
The maximum height of the ball is 12.025 meters.
Example:
Evaluate when and
Solution:
Replace x with 3 in each function  
is undefined.
Example:
Find when
and
Solution:
Example: Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 4; Zeros 23i; 5 multiplicity 2
Solution:
By the Fundamental Theorem of Algebra, since the degree of the polynomial is 4 the polynomial has 4 zeros if you count multiplicity.
There are three given zeros of 23i, 5, 5.
The remaining zero can be found using the Conjugate Pairs Theorem. f(x) is a polynomial with real coefficients. Since 23i is a complex zero of f(x) the conjugate pair of 2+3i is also a zero of f(x).
Now that all the zeros of f(x) are known the polynomial can be formed with the factors that are associated with each zero.
Since f(x) has a zero of 5, f(x) has a factor of x5
Since f(x) has a second zero of 5, f(x) has a second factor of x5
Since f(x) has a factor of 23i, f(x) has a factor of x(23i)
Since f(x) has a factor of 2+3i, f(x) has a factor of x(2+3i)


The polynomial with degree 4 and zeros of 23i and 5 wiht multiplicity 2 is
Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?
Solution: There is a two part process to this problem. Part 1: Use some of the information to find the decay rate of radium. Part 2: Answer the question using the rest of the given information.
Part 1: Find the decay rate of radium.
Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.
Exponential Decay Model
is the initial amount
is the decay rate
is the time
is the amount after t time has passed
Since radium has a half life of 1690 years, we know that whatever initial amount of radium is present after 1690 years there will be half of that initial amount left. This allows me to identify when the .
Substitute these values into the exponential decay formula and solve for k.
Using only the information about radium having a halflife of 1690 years I have found the decay rate for radium.
Note: Although I have put an approximation for k here, try not to round until the very last step.
Part 2: Answer the question using the rest of the given information.
Given information: If 50 grams are present now, how much will be present in 630 years?
With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years. Symbolically that is when the . Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.
, , and 

38.615 grams will be present 630 years later is 50 grams are present initially.
Example: The product of some negative number and 5 less than twice that number is 273. Find the number.
Solution: Translate the statement into an equation and then solve the equation.
“The product of some negative number…”
Let x be some negative number and multiply that number by the expression that comes next in the statement.
“…and 5 less than twice that number…”
5 less than means take 5 away from what follows. Twice that number refers to 2 times the negative number that was described before. (2x5)
“…is 273”
This translates to equals 273.
Solve the equation.
Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation)  
Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula) I will demonstrate factoring so I need to move all terms to the same side to have zero on one side.  
Factor  
Use the zero product property and set each factor equal to zero and solve. 
Since the statement says that the number is negative, the number is .
Hello all! My name is Amanda Sartor and this is my webpage. I use this to provide resources to my in person students and my online students. I don’t have resources for everything but I am always adding. At the top you will find handouts that I use during class and the solutions to the problems. On the left there are blog posts that are specific to a certain problem or problem type. I hope you find these resources helpful.
Amanda Sartor
Example: Find the partial fraction decomposition for the rational expression.
Solution:
Factor the denominator completely  
Write the expression with each factor as a separate fraction.  
Decide what type of expression to put in the numerator
x is linear so we put a constant in the numerator is a quadratic so we put a linear expression in the numerator 

Multiply both sides of the equation by the common denominator and simplify by canceling out common factors  
Use the distributive property and collect the like terms with the x’s  
Equate coefficients to create a system of equations to solve  
On the left 11 is the constant term. There is no linear term so the coefficient is zero and there is no quadratic term so the coefficient is zero.
On the right the constant (no x’s) is 11A, the linear coefficient is C and the quadratic coefficient is A+B. 
Now solve the system of equations.
The first equation is 11=11A. Since it only has one variable, I can solve for A by hand.
The second equation is 0=C . It is already solved.
The third equation is 0=A+B. Since I know A=1, I can substitute and solve for B.
Using the original set up of , substitute the values of A, B, and C.
The partial fraction decomposition of is