There are two different types of forms.

Form One: f(x) =

**ax**

^{2}+ bx + c1. Opens up if a > 0, opens down if a <>

^{2}+ bx + c or use the quadratic formula

Form Two: f(x) =

**a(x-h)**, where a does not equal 0

^{2}+ k1. Vertex (h,k)

2. Axis of symmetry x = h

3. Opens up if a > 0, opens down if a < style="font-weight: bold;">

4. Shift: h units horizontally, k units vertically

Finding Min/Max on calculator:

1. Enter equation into Y1

2. Graph

3. 2nd CALC -> 4: Maximum -> Left, Right, Guess

Monomial and Polynomial Functions:

Degree tells how many roots

If the degree is odd and the leading coefficient is positive -> falls to left, rises to right

If the degree is odd and the leading coefficient is negative -> rises to left, falls to right

If the degree is even and the leading coefficient is positive -> rises in both directions

If the degree is even and the leading coefficient is negative -> falls in both directions

*Any factor ( )

^{2}

^{}is said to have a multiplicity of two

A factor with a multiplicity of two is a "player root". Instead of going through the x-axis like a normal root, "player roots" are tangent to the x-axis.

Polynomial Long Division:

1.Put the polynomial being divided under the division sign and the polynomial being divided by to the left of the division sign.

2.Find how many times the second polynomial fits into the first one by multiplying it and then subtracting that from the first polynomial.

3. If there is no remainder:(a-d are true)

a. The polynomial is a factor.

b. The root(s) are roots of the polynomial.

c. The quotient is also a factor, and the factors of the quotient are as well.

d. The roots of the quotient are roots of the polynomial.

4. If there is a remainder: (e-g are true)

e. The polynomial is not a factor.

f. The root(s) are not roots of the polynomial.

g. f(the root of the dividing polynomial) = the remainder.

5. When dividing a fourth degree polynomial or greater repeat the first two steps with the quotient being the divided polynomial until an answer is reached that cannot be divided further.

example:

Synthetic Division:

Usable when a number is suspected to be a root of an equation to simplify division.

1. The root is put to the left of the upside-down division sign and the coefficients are put to the right and over the division sign. note: if a coefficient is 0, it is included as a placeholder.

2. Begin dividing by taking the first coefficient and bringing it down and then multiplying the root by that coefficient and subtracting that from the second coefficient.

3. This is done until no more coefficients remain.

4. The coefficients are then put back into a polynomial of the degree below it(A polynomial with a variable to the third would become one with a variable to the second and so on)

5. If there is a remainder, the number is not a root of the polynomial and the f(the number divided by) is equal to the remainder.

example:

The Remainder Theorem:

If a polynomial is divided by (x+k) then f(-k) is equal to the remainder.

If a polynomial is divided by (x-k) then f(k) is equal to the remainder.

The Factor Theorem:

A polynomial only has a factor of (x+k) if -k is a root of the equation.

A polynomial only has a factor of (x-k) if k is a root of the equation.

Sample Questions:

1. Find the equation of the parabola whose roots are 3 and -2 and whose y – intercept is 1.

2. Find the equation of the parabola with a vertex is (1,2) and passes through the point (3, - 6).

3. Find the other two roots.

4.

a. What is x³+6x²+15x+6 divided by x+3 equal to?

b. What are all the roots of x³+6x²+15x+6?

5. f(x) = x³+5x²+3x-11, find f(4) using polynomial division or synthetic division.

6. x+3 is a factor of x

^{4}

_{}-x³-11x²+9x+18, find all roots of this equation.

7. The cost of a product is modeled by the equation: C = 800 - 10x + .25x². Find the value for x and c that will minimize cost (Hint: Find Vertex).

8. Describe the end behavior of 2x³+2 and make a sketch.

9. Roots {-1, 0, 0, 3, 5} Write an equation given these roots.

10. Give the degree of:

Answers:

1. y = (1/6)(x

^{2}) - (1/6) x - 1

2. y = -2 (x-1)

^{2}+ 2

3. -2, 1

4. a.) x²+3x+2 b.) -3, -2, -1

5. 145

6. -1, 2, 3, -3

7. x = 20, c = 700

8. Down to left, Up to Right

9. (x+1)(x²)(x-3)(x-5)

10. 5

Selected Answers Explained:

1. Given roots: find factors and multiply

y = (x-3) (x+2)

y = a(x^{2 }- x - 6)

Other information is needed to know which equation, such as a y - intercept (parabola could open up or down)

However, since we know the y - intercept is 1. Just plug in the y - intercept information to find full equation.

y = a(x^{2} - x - 6)

^{2}- 0 - 6)

1 = -6a

a = (-1/6)

y = (-1/6) (x

^{2}- x - 6)

y = (1/6)(x

^{2}) - (1/6) x - 1

2. Easiest to use standard form

*Remember (h,k) is vertex, so...

y = a(x-1)^{2} + 2

Then again, just plug in given point to find a

-6 = a (3-1)^{2} + 2

-6 = 4a + 2

-2 = a

y = -2 (x-1)^{2} + 2

Any Questions?

Feel free to leave a comment.

## 5 comments:

I am not a huge fan of using ^2 for an exponent. It looks a bit lazy.

Make sure you look through the June 2001 exam for items related to this topic. And remember, this isn't a test of whether you can transcribe the notes that are posted on Blackboard. I'm looking for you to distill the essence of what someone needs to know going into the final.

You guys have done a nice job. I like the solution of using a fixed width font and text for the polynomial long division and synthetic division examples.

One thing - the numbering scheme in your polynomial long division procedure is confusing. Consider using letters, or bullets, or something, for the subsections.

Wow this is really easy to understand. I like your formatting!

I was doing number 1 of the practice problems and I got the answer to be (-1/6)x^2+(1/6)x+1 not your answer. and when I tried it in my calc your answer got the y-int to be -1 not 1. So I would check that.

Sorry to be annoying again, but the answer to number 2 is supposed to be the roots- not an equation. Sorry!

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