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## College Algebra

1 review
by: Regina Zauner

25

1

5

# College Algebra MATH 1513

Regina Zauner

GPA 3.5

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Study guide for exam 1. I'm still trying to figure out the graphing part.
COURSE
College Algebra
PROF.
Sabrina Gomez
TYPE
Study Guide
PAGES
5
WORDS
KARMA
50 ?

## 1

1 review
"Better than the professor's notes. I could actually understand what the heck was going on. Will be back for help in this class."
Kaela

## Popular in Mathematics (M)

This 5 page Study Guide was uploaded by Regina Zauner on Thursday February 4, 2016. The Study Guide belongs to MATH 1513 at Oklahoma State University - Oklahoma City taught by Sabrina Gomez in Spring 2016. Since its upload, it has received 25 views. For similar materials see College Algebra in Mathematics (M) at Oklahoma State University - Oklahoma City.

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## Reviews for College Algebra

Better than the professor's notes. I could actually understand what the heck was going on. Will be back for help in this class.

-Kaela

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Date Created: 02/04/16
College Algebra Section 118 Study Guide for test 1 Section 3.4 Perpendicular lines Slope of two lines are m1 and m2, neither line is vertical. −1 The lines are perpendicular if and only if = m2 ) Examples:    Find the equation of the perpendicular line. y= −4 x+2 y= 9x a. 9           a solution:4 b. Line passes through points (­1,3) and (4,1) x2−x1 Solution: since slope =y2−y1  we can find the slope of the original line,  then find the negative reciprocal of the slope of the original line. y= 1−3 4− (1 ) So  m= −2     Then the negative reciprocals would be= 5 .  And the  5 2 5 equation to one perpendicular line would be  = 2x+6 . Determine if the equations are for perpendicular or parallel. a. 3x−7y=12   and   14x+6y=−5 First we need to put both equations into slope intercept form. −7y=−3x+12    and    y=−14x−5 y=−3 x+12 y= −14 x−5 −7                       6  3 12 −7 5 y=7x− 7                    3    6     Now that we have the equations in slope intercept form we can determine if they  are perpendicular or parallel by looking at the slope of each equation.  Remember  lines are perpendicular if and only if their slopes are the negative reciprocals of  each other. 3 −7 So on our first equation the slope 7s   and the second equation’s slope i3 .  These are negative reciprocals of each other so the lines are perpendicular. 5y+1 b. 3− (2y+x =7(x−y)       and      =3+2 x 4 Again place both equations into slope intercept form. 5y+1 3−2y−x=7x−7y                   ( )     (3+2x) 4            y=8x−7y 5y+1=8x+12 −2y=8x−7y−3 5y=8x+11 8x+11 5y=8x−3 y=   5 y=8x−3 y= x+ 11 5 5 5 8 3 y= x− So here we see that both equations have the same slope.   5 5 This means that they are parallel.   Let’s try one more. c. x−4 y=3 4 x−y=2 Again we need to put the equations into slope intercept form. −4y=−x+3 −y=−4x+2 −x+3 y= −4 y=4x−2 y= x− 3 4 4 Now we need to look at the slope for each equation.  Are  the negative reciprocals?  No.  Are they the same slope?  No.   So these equations  are neither parallel nor perpendicular.  Section 3.6 The circle,       √x−h) +(y−k) 2    or     =(x−h) 2 +(y−k) 2 With center (h, k) and radius r. Write the standard form of the equation for the circle. a. Center (­4,­3); radius 5       r =(x­h) +(y­k)    so now we just plug in  numbers into our equation. 2 2 2 2 2  (5) =(x­(­4)) +(y­(­3))     so      25=(x+4) +(y+3) b. Endpoints of a diameter are (0, 6) and (8, 0) Now, we know that the  midpoint of the diameter of a circle is the center.  So, we use the midpoint formula  0+6 8+0 , to find the ordered pair of the center. (  2 2  )    Which is (3, 4).   Now we  can either plug in numbers to find the radius or used the distance formula.  This  time since we have several ordered pairs we will just plug in numbers.   2 2 2 2 2 2 2 R = (0­6) + (8­0)                   R = (­6) + (8)                 R =36+64 R =100                   R=10 So now that we have both the center point and the radius, we can plug the  numbers into the formula to get its standard form.    100=(x­3) +(y­4) 2 2 Section 4.1 a. Find the domain and range of the following set of ordered pairs.   {(−2,5 ),−2,3 ),−2,0 ),(−2,9) } R=   the domain is ­2 and the range is (5, 3, 0, 9) Now if we graphed this relation would it be a function?    No,6 because each x  value has multiple Y values. −3 )−5+5x ) What is the implied domain of the following:  h(x) =    Domain = x {x ≠0}   b. Find the difference quotient {[f(x+h)− f(x]}        x)=−3x +2x+6 h Now we substitute in the value of f(x) in the equation and in the difference  [−3 x+h )+2 x+h +6 ] −3x +2x+6 )} part we put (x+h). ⟨ h ⟩ −3(x+h )x+h +2 x+h )+6− −3x +2x+6 ) ⟨[ }⟩ h 2 2 2 [−3 x +2 xh+h )+2 x+2 h+6+3 x −2 x−6] { h } −3x −6xh−3h +2x+2h+6+3x −2x−6 ) h −3h −6xh+2h ) h(−3h−6x+2 ) h [ h ] −6x−3h+2 This type of problem was giving me lots of problems because I was  substituting wrong.  Make sure you use the entire equation in both parts of  the numerator but where the (x+h) part is you put (x+h) instead of just x. Section 4.2

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