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by: Kait Brown

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# Chapter 2.0-2.3 MATH 1100.140

Kait Brown
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Everything in Chapter two thus far
COURSE
College Algebra
PROF.
Mary Ann Barber
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Graphs, functions, equations, symmetry, Lines, x-intercepts, y-intercepts
KARMA
25 ?

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This 6 page Class Notes was uploaded by Kait Brown on Monday October 3, 2016. The Class Notes belongs to MATH 1100.140 at University of North Texas taught by Mary Ann Barber in Fall 2016. Since its upload, it has received 4 views. For similar materials see College Algebra in Math at University of North Texas.

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Date Created: 10/03/16
Chapter 2: Graphs and Functions 2.1.1. Place the following points on the axis: A (2,-3) B (-3,4) C (0,5) D (-6,0) E (-4,-7). 6 C B 4 2 0 D -7 -6 -5 -4 -3 -2 -1 0 1 2 3 -2 A -4 -6 E -8 2.1.2. What does the Pythagorean Theorem say? What does this say about the distance between the points (x ,y ) 1 1 and (x2,y2) in the picture below? 2 2 2 3.5 ???? + ???? = ???? √ 2 2 3 (x 2y2) ???? = ???? + ???? ???? = √ (????2− ???? 1 + (???? − 2 ) 1 2 2.5 2 1.5 1 (x ,y ) 1 1 0.5 0 0 0.5 1 1.5 2 2.5 + 2.1.3. What point is halfway between (2,8) and (6,0)? Add x’s and divide by 2. Add y’s and divide by 2. Write as ordered pair. 2+6 = 4 0+8 ) = 4 (4,4) 2 2 2.1.4. Find the distance between the points (1,3) and (-2,7). ???? = √ (−2 − 1) + (7 − 3) = −3 + 4√= 9 + 16 = 25 =√5 √ 2.1.5. Find x so that the point (x,8) is 13 units from (3,-4). ???? = 13 13 = 2 √ (3 − ????) + (−4 − 8) 2 169 = √ (3 − ????) + 144 169 − 144 = 25 3 − ???? = 5 ???? = 3,−2 2.1.6. The midpoint of the line segment from P t1 P is2(5,-4). If P 2(7,-2), what is P1? (5,−4) = (????,????) − (7,−2)3 4 5 6 7 5 is the midpoint of 3 and 7. -6 -5 -4 -3 -2 -4 is the midpoint of -6 and -2. So (x,y)=(3,-6) (x,y)=(3,-6) 2.2. Graphs of Equations Some examples of graphs. y=3x-2 X y 6 -2 -8 -1 -5 4 0 -2 1 1 2 2 4 An equation 0 whose -3 -2 -1 0 1 2 3 variable has a -2 coefficient (like 3x) -4 denotes a straight line. -6 -8 -10 2 y=x +1 6 x y -2 5 5 -1 2 0 1 1 2 4 2 5 An equation whose 3 variable is raised to a power (like x ) denotes a parabola. 2 1 0 -3 -2 -1 0 1 2 3 4 x y -2 -8 -1 -5 3 0 -2 1 1 2 2 4 *Lacks sufficient data to 1 determine shape 0 -2 0 2 4 6 8 10 12 -1 -2 -3 Intercepts  To find the x-intercepts, set y=0 and solve for x.  To find the y-intercept, set x=0 and solve for x. 2 2.2.1. Find the x- and y-intercepts of y=x -x-6 ???? − ???????????????????????????????????????? → ????ℎ???????????????? ???????????? ???? ???? ???????? 0 → 0 = ???? − ???? − 6 → 0 = ???? − 3 ???? + 2 ???? = 3,−2 ???? − ???????????????????????????????????????? = (3,0 ,(−2,0) ???? − ???????????????????????????????????????? → ????ℎ???????????????? ???????????? ????^′ ???? ???????? 0 → ???? = 0^2 − 0 − 6 → ???? = −6 ???? − ???????????????????????????????????? = (0,−6) 2.2.2. Draw a graph that is symmetric across the x-axis from the graph below 10 8 6 4 2 0 -2.5 -2 -1.5 -1 -0.5 0 -2 -4 -6 -8 -10 2.2.3. Draw a graph that is symmetric across the y-axis from the graph below. 3 2 1 0 -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -3 -4 -5 -6 -7 Test for symmetries  A graph is symmetric with respect to the y-axis if, for every point (x,y) on the graph, the point (-x,y) is also 2 on the graph. Consider y=x . Substituting –x for x in the equation yields the original equation.  A graph is symmetric with respect to the x-axis if, for every point (x,y) on the graph, the point (x,-y) is also 2 on the graph. Consider x=y . Substituting –y for y in the equation yields the original equation.  A graph is symmetric with respect to the origin if, for every point (x,y) on the graph, the point (-x,-y) is also 3 on the graph. Consider y=x . Substituting –x for x and –y for y in the equation yields the original equation. A circle is a set of points in the plane that are at a fixed distance, r, from a specified point (h,k). The fixed distance r is called the radius and the specified point (h,k) is called the center point. Standard Form for the Equation of a Circle 2 2 2 The equation of a circle with center (h,k) and radius r is (???? − ℎ) + (???? − ????) = ???? and called the standard form of an equation of a circle. Note that is (h,k)=(0,0) and r=1, then (x-0) +(y-0) =1 can be written more simply as 2 2 x +y =1. This special circle is called the unit circle. 5 7 2.2.4. Determine the types of symmetry for the graph of y=3x +7x . ???????????????????????????????? ???????? ???? − ???????????????? ???????????????????????????????? ???????? ???? − ???????????????? ???????????????????????????????? ???????? ???????????????????????? 5 7 5 7 5 7 ???? ≠ 3(−????) + 7(−????) − ???? ≠ 3???? + 7???? − ???? = 3(−????) + 7 −???? ( ) ???? = 3???? + 7???? 7 2 2 2.2.5. Find the center and radius of the circle given by the equation x -6x+y +10y=0. ???? − 6???? + ___________ + ???? + 10???? + __________ = 0 + ___ + ____ Coefficient:-6 Coefficient: 10 ½ coefficient: -3 ½ coefficient: 5 2 2 Coefficient : 9 Coefficient : 25 ???? − 6???? + 9 + ???? + 10???? + 25 = 0 + 9 + 25 (x-3)(x-3) (y+5)(y+5) 2 2 (x-3) +(y+5) =34 Center: (3,-5) Radius: √???? 2.3 Lines Slope of a Line The slope of a non-vertical line that passes through the points P(x ,y ) and Q(x ,y ) is denoted m and is defined by 1 1 2 2 ???? = ???? 2???? 1. Rate of change.∆???????????????? ∆???? ????2−????1 = ???? ???? 2???? 1 ∆???????????? ∆???? ????2−????1 The slope of a vertical line is undefined. Types of equations of a line:  Point-Slope: ???? − ????1= ????(???? − ???? ) 1  Slope-Intercept:???? = ???????? + ????  Standard:???????? + ???????? = ???? 2.3.1. Sketch the graphs of the following lines: 3 ???? − 4 = 2 ???? + 1 ) ???? = − ???? + 2 ???? = 3 ???? = −2 7 ???? = 2 ???? = − 3 ???? = 0 ???? = ???????????????????????????????????? 1 7 Positive (m>0) Negative (m<0) 0 (m=0) no slope Parallel and Perpendicular Lines 2.3.2. What is an equation of the line through the point (1,7) and parallel to the line passing through the points (2,5) and (-2,1)? 1−5 −4 Point: (1,7) Slope (of parallel line using points gi−2−2:= −4 = 1 ???? − ????1= ???? ???? − ???? 1)???? − 7 = 1 ???? − 1 ???? = ???? − 1 + 7 ???? = ???? + ???? 2.3.3. Find the equations of the vertical and horizontal lines though the point (-3,5). Vertical line: x=-3 Horizontal line: y=5 Vertical and Horizontal Lines  A vertical line crossing through the point (a,b) has an equation x=a  A horizontal line crossing through the point (a,b) has an equation y=b Simple linear regression is an approach to modeling the relationship between a dependent variable y and an explanatory variable x. The line generated by this method is called the regression line or the least squares line. 2.3.4. The equation of a line is given as 2x+5y=-6. Find the x-intercept and the y-intercept then use them to graph the line. *Note that -3=-6/2 x 0 -3* y -6/5 0 2.3.5. Use the following points to find a regression line and predict the y values for x values 4 and 6. (1,3), (2,5), (3,6), (4,10), (5,11). In your calculator y=21x+.7 1. Stat 3. 2 Quit 5. Lin. Reg. 2. Edit 4. Stat ͢ Calculate 6. Hit enter until you see an equation, then write it down. 2.3.6. What is the y-intercept of the line passing through the points (-2,6) and (4,-3)? ???? = ????2−????1 −3−6= =9 −3 ???? − 6 = − 3 (????—2 ) ???? = 0 ͢ (???? − 6 = − 3(0 + 2 = −3 y=3 ????2−????1 4—−2 6 2 2 2 2.3.7. Determine the value of the x-coordinate of the point (h,5) that lies on the line y=3x-7. 12 3ℎ 5 = 3ℎ − 7 12 = 3ℎ 3 = 3 4 = ℎ ???? − ???????????????????????????????????????? = 4 2.3.8. Find an equation of the line perpendicular to 5x-10y=11 that passes through the point (1,-1). −10???? −5???? 11 1 11 1 2 −10???? = −5???? + 11 −10 = −10 + −10 ???? = ????2− 10 So if ???? = 2⊥ ???? = − 1 ???? − ???? = ???? ???? − ???? ) ????—1﷧ = −2(???? − 1) ???? + 1 = −2???? + 2 ???? = −2???? + 1 1 1 2.3.9. Find the equations of the vertical and horizontal lines through the point (5,-11). Vertical line: x=5 Horizontal line: y=-11

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