MATH 171 Final Review

find the vertex of a parabola when given in f(x)=ax²+bx+c

(-b/2a,f(-b/2a))

Graph exponential functions: y-int of f(x)=bˣ

(0,1)

Standard form of a circle

(x-h)²+(y-k)²=r² The center is (h,k)

x⁰

1

Solve Rational Inequalities

1. Put into the form of the function on one side and 0 on the other 2. Find the x-intercept(s) by setting the numerator of the rational function to 0 and solving 3. Find where the function is undefined by setting the denominator equal to 0 and solving 4. The x-int(s) and the number(s) where the rational function is undefined are the boundary points. 5. Plot the boundary points on a number line to separate the line into intervals. 6. Choose a test number within each interval, for each interval, and evaluate f at that number) if the test number makes the inequality true then that range is included in the solution set) if the test number makes the inequality false then that range isn't included in the solution set If the original inequality is < or > then don't include the boundary points in the answer. If the original inequality is ≤ or ≥ then include the ZERO boundary points in the answer, but not the boundary points from the denominator being undefined

Piecewise function

2 or more equations over a specific domain Graph by plotting points in each equation's domain. Remember that you are given the y value IF the x value meets a certain criteria. If your graph looks weird make sure they follow the pattern of whatever graph the equation would make (quadratic, linear, etc.)

Composition of Functions

A function is performed, and then a second function is performed on the result of the first function. (f○g)(x) = f(g(x)). f(g(x)) means that function g is the input to function f. Wherever x is in f(g(x)) replace it with (g(x)). and then wherever (g(x)) is in f(g(x)) replace it with g(x) Domain: the set of all x such that a) x is in the domain of g b) g(x) is in the domain of f Ex. f(x) = 5x+6; g(x) = 2x²-x-1; find (f○g)(x) 5(g(x))+6 → 5(2x²-x-1)+6 → 10x²-5x-5+6 → 10x²-5x+1 D𝒻₍ₓ₎ = (-∞,∞) because there are no denominators or roots D𝓰₍ₓ₎ = (-∞,∞) because there are no denominators or roots. So the domain is (-∞,∞) Ex. f(x)= 5x+6; g(x) 2x²-x-1; find (f○g)(-1) 5(g(-1))+6 → 5(2(-1)²-(-1)-1)+6 → 5(2+1-1)+6 → 5(2)+6 → 10+6 → 16. The domain is (-∞,∞)

one-to-one function

A function where each element of the range is paired with exactly one element of the domain. Ex. f(x)=x+1 different inputs will never receive the same answer. Are continually increasing or decreasing but not both Pass a horizontal line test with only one spot

one to one function

A function where each element of the range is paired with exactly one element of the domain. Ex. f(x)=x+1 different inputs will never receive the same answer. Are continually increasing or decreasing but not both (per line.) Pass a horizontal line test with only one spot

asymptote

A line that a graph approaches but doesn't cross The line x=a is a vertical asymptote if as x approaches a, f(x) increases or decreases without bound. So as x approaches a from left or right, f(x)→∞ or f(x)→-∞ The line y=b is a horizontal asymptote if as f(x) approaches b, x increases or decreases without bound As different translations are applied to the function, its asymptote moves too

polynomial function

A smooth continuous (unbroken) curve. Domain is (-∞,∞) linear quadratic cubic

Ways to factor

AC method grouping Quadratic Formula possible rational zeros rinse and repeat

Graph exponential functions: The range of f(x)=bˣ

All positive real numbers (0,∞)

Graph exponential functions: The domain of f(x)=bˣ

All real numbers (-∞,∞)

Graph exponential functions: f(x)=bˣ passes through the point (0,1)

Because f(0)=b⁰=1 (b ≠ 0)

1/x²

Common rational function Even function f(x)=f(-x) Y-axis symmetry Domain (-∞,0)∪(0,∞) OR x≠0 *0 is where the function is undefined*

1/x

Common rational function Odd function f(x)=-f(x) Origin symmetry Domain (-∞,0)∪(0,∞) OR x≠0 *0 is where the function is undefined*

fractions

Don't make them mixed numbers. They are fully simplified as improper fractions

Find zeros of a 4th degree polynomial function missing terms

Ex. f(x)=x⁴-4x²-32 1. Factor - Comparing the possible rational zeros to the graph on the calculator there are no rational zeros. Can't factor this way. - Can't factor with the quadratic formula since the equation is of degree 4 not 2. - Can't factor out 4 or x² - The AC method will work though since we are working with 3 terms. 4 and -8 work. x⁴-8x²+4x²-32 Now factor out x² x²(x²-8)+4(x²-8) Factor out (x²-8) (x²+4)(x²-8) 2. Use zero factor theorem to solve x²+4=0 x²=-4 x=±2i And x²-8=0 x²=8 x=±2√2 3. Your solutions are x=±2i, ±2√2

Find a polynomial frunction from its zeros: expressional roots of an imaginary number

Ex. x=-2+5i, x=1 Imaginary expressions come in conjugate pairs so -2-5i is also a zero of the function

Find a polynomial frunction from its zeros: fractional zero

Ex. x=3, x=-5, x=3/4 f(x)=(x-3)(x+5)(x-3/4) Since we need *integer* coefficients we have to get rid of the denominator. This can be done by multiplying x in the relevant factor by the denominator. f(x)=(x-3)(x+5)(4x-3)

Find a polynomial frunction from its zeros: expressional roots of a constant

Ex. x=4±√3, x=0 Since one zero is 0 that means that it is just x by itself f(x)=x(x-(4+√3))(x-(4-√3)) Distributing the negative just switches the factors around. f(x)=x(x-4-√3)(x-4+√3) Use the property that states (a-b)(a+b)=a²-b² with x-4 being a and √3 being b f(x)=x((x-4)²-(√3)²) Simplified f(x)=x((x-4)²-3) Use the principle that (a-b)²=a²-2ab+b² f(x)=x(x²-8x+16-3) Distribute f(x)=x³-8x²+13x

Find the zeros of a 4th degree polynomial function with all terms

Ex. x⁴-10x³+34x²-42x+9 1. Factor - Comparing the possible rational zeros to the graph on the calculator 3 could be a zero of the equation Use synthetic division to test 3⟌1,-10,34,-42,9 *Remember to include a 0 for any missing degrees* Fully "divided" it becomes 1,-7,13,-3,0 Since we got no remainder 3 *is* a zero of x Factor as 0=(x-3)(x³-7x²+13x-3) 2. Factor again - We can't factor by grouping - The quadratic formula won't work since it's of degree 3 not 2 - Nor the AC method - Try the possible rational zero theorem again The poss. rational zeros ± 1 and ±3 Graphing, 3 could be a zero Test with synthetic division 3⟌1,-7,13,-3 Fully "divided" it becomes 1,-4,1,0 Since we got no remainder 3 *is* a zero of x Factor as 0=(x-3)²(x²-4x+1) 3. Factor again again - We can't use the AC method - We CAN use the quadratic formula [4±√(16-4)]/2 2±(2√3)/2 2±√3 4. The Answers x=3 with a multiplicity of 2 x=2+√3 x=2-√3

Rational Zero Theorem

Find all *possible* rational zeros p is all the factors for the constant term. q is all of the factors of the leading coefficient. If a polynomial has integer coefficients, and p/q is a rational zero of f, then p is a factor of the constant term a₀, and q is a factor of the leading coefficient, aₙ Each of its rational zeros has the form p/q Find p, the factors of the constant term, and divide that by q, the factors of the leading coefficient. Divide each number in the numerator by each number in the denominator, those are your possible rational zeros/roots

Solve absolute value inequalities

If |x| < a then -a<x<a If |x| ≤ a then -a≤x≤a If |x| > a then x > a or x < -a (Use the union sign for interval notation) If |x| ≥ a then x ≥ a or x ≤ -a (Use the union sign for interval notation) If |x| </≤ -a then there is no solution If |x| >/≥ -a then all real numbers are solutions

Given a graph sketch the graph of another function which involves the original function's graph

Just apply the transformations to the graph, no need to plot the points remember the order: horizontal, scaling, reflections, then vertical

Find a polynomial with integer coefficients when given it's zeros

Linear factorization theorem - The factors of a function are 0=(x-c₁)(x-c₂)(x-cₑₜ𝒸) where c is the zero.

Test a rational zero

Not all rational zeros will be real zero solutions. But all real zero solutions will be in the rational zero test. The rational zero test only limits the number of possible zeros 1. Graph on the calculator, then verify zeros through synthetic division or the quadratic formula or some other form of factoring. Find the rational possible zero you want to test and make that the divisor in synthetic division. 2. Getting a 0 remainder means that the divisor is a real zero of the function. Getting a remainder means it is not a real zero, only a possible one 3. Further, factor out the quotient and solve using the zero factor theorem. You are now finding the solutions for zero so the answers you find with the zero factor theorem are included in the set, not excluded. Finding all the zeros includes trial and error. Once you find one of the zeros though, you can factor the whole equation and use the zero factor theorem to find the rest

How to tell a function from a graph

Vert line test. If anywhere on the graph a vertical line crosses more than one point then it is not a function

How to tell a function from an equation

When fully simplified if x is raised to a higher power that's fine but if y is raised higher than 1 then it is not a function

divide polynomials

When the degrees of the numerator is 1 degree larger than the denominator divide the numerator by the denominator Ex. 2x²-x-6/2x-3 1. Can't use synthetic division because it is not in the form x-c. Use long division instead 2x-3⟌2x²-x-6 How many times does 2x go into 2x²? x times. Multiply that by the divisor and then subtract that from the dividend 2x²-x - (2x²-3x) = 2x How many times does 2x go into 2x? 1. Multiply that by the divisor and then subtract that from the dividend 2x-6 - (2x-3) = -3 The answer is x+1+[3/(2x-3)] So the slant asymptote is y=x+1

synthetic division

a method used to divide a polynomial by a binomial (x-c) 1. Arrange the polynomial by their descending degrees. Tack on a 0 coefficient for any missing degrees terms including 0 degrees (constants) 2. write c for the divisor, x-c (not -c). 3. To the right, list only the coefficients of the dividend 4. Bring down the first coefficient of the dividend to under the line 5. Multiply c by the value just written on the bottom row. Write the product in the next column in the 2nd row 6. Add the values in this new column. Write the sum in the bottom row 7. Repeat this multiplying and adding till all the columns are filled in 8. Use the numbers in the last row to write the quotient plus the remainder above the divisor. The degree on the 1st term of the quotient is 1 less than the degree of the dividend. The final value in this row is the remainder Synthetic division doesn't actually include any division. Just multiplying and adding

rational zero

a zero (solution) of a function which can be written as p/q

intercepts

answer as point(s)

(a+b)²

a²+2ab+b²

(a-b)²

a²-2ab+b²

(a-b)(a+b)

a²-b²

locate horizontal asymptotes

f(x) = (aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀)/(bₘxᵐ + bₘ₋₁xᵐ⁻¹ + ... + b₁x + b₀) The degree of the numerator is n The degree of the denominator is m There is a term for each degree, even if the term is 0. 1. if n < m, then y = 0 (the x-axis) is the horizontal asymptote of f(x) 2. if n = m, then the line y = aₙ/bₘ is the horizontal asymptote of f(x) 3. if n > m, then there are no horizontal asymptotes of f(x) Horizontal asymptotes can be crossed in the middle but will never be crossed near the end behaviors Ex. 9x²/3x²+1 n = m so y = aₙ/bₘ y = 9/3 y = 3 is the horizontal asymptote of f(x)

strategy for graphing rational functions

f(x) = p(x)/q(x) Where p and q are polynomial functions and fully simplified 1. Find if there is symmetry (f(x)=f(-x) is y-axis symmetry. f(x)=-f(x) is origin symmetry) 2. Find the y-int (if there is one) by evaluating f(0) 3. Solve for the x-ints 4. The 2 "pieces" are separated by the vertical asymptote(s), test a point within every region separated by the vertical asymptotes and that will show whether the "piece" is above or below the horizontal asymptote 5. Really you just have to plug numbers and solve then graph. You can also refer to the graph on your graphing calculator for the generl shape but beware since it graphs the asymptotes weird and it's hard to read exact numbers from a graph 6. DON'T FORGET THE ASYMPTOTES AND THAT THEY ARE DOTTED

Graph exponential functions: b > 1

f(x)=bˣ has a graph that goes up to the right and is an increasing function. The greater the value of b, the steeper the increase

transformations

f(x)=x+c shift upwards f(x)=x-c shift downwards f(x)=(x+c) shift left f(x)=(x-c) shift right f(x)=-x reflect about x-axis f(x)=(-x) reflect about y-axis Order to perform transformations 1. Horizontal shifting 2. Stretching or shrinking 3. Reflections, doesn't matter which (x-axis/y-axis) 4. Vertical shifting

Cubic Function

f(x)=x³ Domain (-∞,∞) Range (-∞,∞)

Absolute Value Function

f(x)=|x| Domain (-∞,∞) Ranges [0.∞)

Square Root function

f(x)=√x Domain [0, ∞) Range [0, ∞)

odd function

graph is symmetrical with respect to the origin; f(-x)=-f(x)

even function

graph is symmetrical with respect to the y-axis; f(x) = f(-x)

locate vertical asymptotes

if f(x)=p(x)/q(x) is a rational function in which p(x) and q(x) share no common factors (so its fully simplified) and a is a zero of q(x), then x=a is a vertical asymptote of f(x). Not all rational functions have asymptotes Vertical asymptotes can NEVER be crossed Ex. Find the vertical asymptotes of f(x)=x/(x²-1) Find the zero of q(x) x²-1 = 0 (x-1)(x+1) = 0 x = 1, -1 are the vertical asymptotes of f(x) Ex. Find the vertical asymptotes of f(x)=x-1/(x²+1) Already fully simplified f(x) has no vertical asymptotes

Graph exponential functions: x-int of f(x)=bˣ

no x-int cause b ≠ 0.

Subtracting numbers from the outside of the fractions

shifts the graph downwards

Adding numbers to the denominator inside the power

shifts the graph left

Subtracting numbers from the denominator inside the power

shifts the graph right

Adding numbers to the outside of the fraction

shifts the graph upwards

numbers in the numerator

stretch the graph

negative leading coefficient w/even multiplicity

the graph falls to the left and right

positive leading coefficient w/odd multiplicity

the graph falls to the left and rises to the right

positive leading coefficient w/even multiplicity

the graph rises to the left and right

negative leading coefficient w/odd multiplicity

the graph rises to the left falls to the right

minimum or maximum

x value gives the location, or "where", of the value y value gives that value x=-b/2a (it's the vertex's x-value) y=f(-b/2a) (it's the vertex's y-value)

x falls to the left

x→-∞, y→-∞

x rises to the left

x→-∞, y→∞

x falls to the right

x→∞, y→-∞

x rises to the right

x→∞, y→∞

exponential function

y=abˣ b is a positive constant other than 1 (b > 0 and b ≠ 1) and x is any real number Ex: f(x)=2ˣ Ex: f(x)=10ˣ⁺¹ Ex: f(x)=(3/4)^(ˣ/³) Not ex: f(x)=xˣ Not ex: f(x)=1ˣ Not ex: f(x)=(-3)ˣ Not ex: x³

quadratic function

y=x² A function in which the greatest power of the variable is 2. Domain (-∞,∞) Range [0,∞)