MAT-261: Pre-Calculus

MAT-261: Pre-Calculus

MAT ­261
Section 1.2 – Visualizing & Graphing Data
Cartesian Coordinate System
Ordered pairs= (x, y)
Domain: set of all possible input values
Range: set of all possible output values
Pythagorean Theorem
Distance Formula
d= √ ( ( x2−x1)
+( y2−y1)
 The set of all points in a plane are an equal distance from the center point
 That distance is the radius (r)
 When centered at (0,0) use: x²+y²=r²
 When centered at (h,k) use: (x­h)²+(y­k)²=r²

o Square root of a negative
When finding the range:
o Use a graph!
Section 1.4: Types of Functions & Their Rates of Change
Slope represents the average rate of change of y with respect to x
M= rise/run
The x­ intercept (horizontal intercept) is the point where the graph crosses the x­axis
The y­ intercept (vertical intercept) is the point where the graph crosses the y­axis
Slope­intercept form is y=mx+b
M is slope
(0, b) is y­intercept
Difference Quotients
Average rate of change (similar to slope of a line)

f ( x+h )− f (x )
Section 2.1: Equations of Lines
Two equations for a line
o Slope intercept form: y=mx+b
o Point slope form: y−y1=m(x−x1
o M is slope in both equation



Section 3.1: Quadratic Functions & Models
 The simplest quadratic function
o F(x)=x²
o Vertex at (0,0)
o Symmetry at x=0
o Decreasing on (-∞,0)
o Increasing on (0, ∞)
 Vertex form of a quadratic
o F(x)= a (x-h) ²+k
o Vertex at (h, k)
o If a>0, the vertex is the minimum
o If a<0, the vertex is the maximum
o Compare to expanded form f(x)= ax²+bx+c
 Domain and Range
o Domain is the all the x values that work
 Domain of any quadratic is (-∞, ∞)
o Range is all the y values that the function hits
 If a>0, start at the vertex and go up forever.
 If a<0, start down at -∞ and go up to the vertex
Section 3.2: Quadratic Equations & Problem Solving
 The Zero Product Property
o If a × b =0, then a=0, b=0, or both
 A second degree or quadratic equation is an equation that can be written in
the form ax²+bx+c, where a, b, and c are real numbers and a is not zero.
 Y-intercept (vertical intercept)
o Crosses the y-axis at (0, c)
 X-intercepts (horizontal intercept)
o For any function solve f(x)=0
o For a quadratic function
 Factor
 Complete the square
 Use the quadratic formula
 The Discriminant
o b²-4ac
o Determines number and type of solutions
Section 3.3: Complex Numbers
 Imaginary i
o √-1= i
o i² = -1
o The standard form for an imaginary number is z=a+bi
o To add or subtract imaginary numbers, you combine like term

MAT-261: Pre-Calculus

week 3

Section 4.2: Polynomials Functions and Models
 Names of polynomial functions
o Degree 2
o Degree 3
o Degree 4
 Number of turning points
o The number of turning points of the graph of a polynomial function of
degree n ≥ 1 is at most n-1
o If a polynomial has t > 0 turning points, its degree must be at least t+1
 End Behavior
o Highest degree term “wins” for large x and negative x
 Zeros of a function
o For real numbers, these are all the same:
 X-intercepts of a graph
 Zeros of a function
 Roots of a polynomial function
 Solutions of p(x)= 0
o If the solution/roots/zeros have a nonzero, imaginary part, there are no
 Even and Odd Functions
o Even function: f(-x) = f(x)
 Symmetric about the y-axis
 If (x, y) is on the graph, then (-x, y) is too
o Odd function: f(-x) = -f(x)
 Symmetric about the origin
 If (x, y) is on the graph, then (-x,-y) is too
o Even functions
 Polynomials with only even powers
 Cos(x)
 Sec(x)
o Odd functions
 Polynomial with only odd powers
 Sin(x)
 Tan(x)
 Csc(x)
 Cot(x)
o Both even and odd f(x)=0
 High and Low points
o Extremum: maximum or minimum
o Extrema: minima or maxima
o Local or relative extrema
 Highest or lowest point in the neighborhood
 Occurs at the turning point of polynomials
o Absolute maximum


week 4: quiz


Ch. 5
1. (i) limx1x21
x+1 = limx1(x1) = 0.
(ii) limx2x38
x2= limx2(x2)(x2+2x+4)
x2= limx2(x2+ 2x+ 4) = 12.
(iii) limx3x38
1= 19.
(iv) limxyxnyn
xy= limxy(xn1+xn2y+···+x2yn3+xyn2+yn1) =
(v) limyxxnyn
(vi) limh0a+ha
h= limh0(a+ha)(a+h+a)
h(a+h+a)= limh0a+hh
= limh01
a+h+a= 1/2a.
2. (i) limx11x
1x= limx11x
(1x)(1+x)= limx11
1+x= 1/2.
(ii) limx011x2
x= limx0(11x2)(1+1x2)
x(1+1x2)= limx01(1x2)
= limx0x
1+1x2= 0/2 = 0.
(iii) Similar to (ii), except you get = limx01
1+1x2= 1/2 at the en
MAT-261: Pre-Calculus click here for help
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