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MAT-261: Pre-Calculus
MAT ­261Section 1.2 – Visualizing & Graphing DataCartesian Coordinate SystemOrdered pairs= (x, y)Domain: set of all possible input valuesRange: set of all possible output valuesPythagorean Theoremlog²+log²=hypotenuse²Distance Formulad= √ ( ( x2−x1)2+( y2−y1)2Circles 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 negativeWhen finding the range:o Use a graph!Section 1.4: Types of Functions & Their Rates of ChangeSlopeSlope represents the average rate of change of y with respect to xM= rise/runInterceptsThe x­ intercept (horizontal intercept) is the point where the graph crosses the x­axisThe y­ intercept (vertical intercept) is the point where the graph crosses the y­axisSlope­intercept form is y=mx+bM is slope(0, b) is y­interceptDifference QuotientsAverage rate of change (similar to slope of a line)
f ( x+h )− f (x )hSection 2.1: Equations of LinesTwo equations for a lineo Slope intercept form: y=mx+bo Point slope form: y−y1=m(x−x1)o M is slope in both equation

WEEK 2:
Section 3.1: Quadratic Functions & Models The simplest quadratic functiono F(x)=x²o Vertex at (0,0)o Symmetry at x=0o Decreasing on (-∞,0)o Increasing on (0, ∞) Vertex form of a quadratico F(x)= a (x-h) ²+ko Vertex at (h, k)o If a>0, the vertex is the minimumo If a<0, the vertex is the maximumo Compare to expanded form f(x)= ax²+bx+c Domain and Rangeo 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 vertexSection 3.2: Quadratic Equations & Problem Solving The Zero Product Propertyo If a × b =0, then a=0, b=0, or both A second degree or quadratic equation is an equation that can be written inthe 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)=0o For a quadratic function Factor Complete the square Use the quadratic formula The Discriminanto b²-4aco Determines number and type of solutionsSection 3.3: Complex Numbers Imaginary io √-1= io i² = -1o The standard form for an imaginary number is z=a+bio 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 functionso Degree 2o Degree 3o Degree 4 Number of turning pointso The number of turning points of the graph of a polynomial function ofdegree n ≥ 1 is at most n-1o If a polynomial has t > 0 turning points, its degree must be at least t+1 End Behavioro Highest degree term “wins” for large x and negative x Zeros of a functiono 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)= 0o If the solution/roots/zeros have a nonzero, imaginary part, there are nox-intercepts Even and Odd Functionso Even function: f(-x) = f(x) Symmetric about the y-axis If (x, y) is on the graph, then (-x, y) is tooo Odd function: f(-x) = -f(x) Symmetric about the origin If (x, y) is on the graph, then (-x,-y) is tooo 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 pointso Extremum: maximum or minimumo Extrema: minima or maximao Local or relative extrema Highest or lowest point in the neighborhood Occurs at the turning point of polynomialso Absolute maximum

week 4: quiz

Ch. 5
1. (i) limx→1×2−1
x+1 = limx→1(x−1) = 0.
(ii) limx→2×3−8
x−2= limx→2(x−2)(x2+2x+4)
x−2= limx→2(x2+ 2x+ 4) = 12.
(iii) limx→3×3−8
x−2=19
1= 19.
(iv) limx→yxn−yn
x−y= limx→y(xn−1+xn−2y+···+x2yn−3+xyn−2+yn−1) =
nyn−1.
(v) limy→xxn−yn
x−y=nxn−1.
(vi) limh→0√a+h−√a
h= limh→0(√a+h−√a)(√a+h+√a)
h(√a+h+√a)= limh→0a+h−h
h(√a+h+√a)
= limh→01
√a+h+√a= 1/2√a.
2. (i) limx→11−√x
1−x= limx→11−√x
(1−√x)(1+√x)= limx→11
1+√x= 1/2.
(ii) limx→01−√1−x2
x= limx→0(1−√1−x2)(1+√1−x2)
x(1+√1−x2)= limx→01−(1−x2)
x(1+√1−x2)
= limx→0x
1+√1−x2= 0/2 = 0.
(iii) Similar to (ii), except you get = limx→01
1+√1−x2= 1/2 at the en