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# MATH121, Unit 2 Test Study Guide Math 121

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This 7 page Study Guide was uploaded by Mallory McClurg on Friday September 30, 2016. The Study Guide belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 33 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/30/16

MATH121 Un niit Teesst 2 Leesssoonnss 3..2,,3..3,, 3.4,, 4..1,,4..5,,& 44.6 We need to be able to… • Find the intercepts of an equation and graph. • Find the slope of an equation given 2 points. • Write the slope intercept form of the equation of a line given 2 points. • Find the slope, y-intercept, and graph given the equation of a line. • Convert a given equation to slope intercept form. • Find the equation of a line which is parallel and/or perpendicular do a given line. • Find the implied domain of a given function. • Evaluate functions (including at algebraic expressions). • Find the difference quotient of a given function. • Algebraically combine functions & find domains. • Compose functions. • Find the inverse of a given function. EXAMPLE 1. Find possible x and y-intercepts, then graph the equation. 11x + 8 = -11y + 8 (First, get y on a side by itself.) 11x = -11y x = -y y = -x (Since one of the variables is negative, there will be a negative slope that runs through (0, 0). Note: positive slopes start lower on the left, and go up towards the right. Negative slopes start higher on the left, and go down towards the right. ) EXAMPLE 2. Find g(x – 1) when g(x) = -6x +4x. 2 (The first thing we need to do here is substitute (x – 1) everywhere we see an x in the g(x) function.) g(x – 1) = -6(x – 1) + 4(x – 1) (First FOIL out the (x – 1) . Then distribute the -6 to the resulting quadratic and then distribute the 4 to the x and -1.) g(x – 1) = -6x + 12x – 6 + 4x – 4 (Combine like terms.) 2 g(x – 1) = -6x + 16x – 10 (This is our answer!) EXAMPLE 3. Find the slope and the y-intercept of this linear equation, then simplify. -5x + 4y = 16 (To find the slope, first, you need to get y on a side by itself.) 4y = 5x + 16 (Divide everything by 4.) y = (5/4)x + 4 (The slope is (5/4). Now, to find the y intercept, let x = 0.) y = (5/4)(0) + 4 MATH121 Unni it Teesstt 2 Le esssoonnss 3..2,, 3..3,, 3..4,, 4..1,, 4..5,, & 4..6 y = 4 y-intercept: (0, 4) (This is our answer!) EXAMPLE 4. Put this equation into slope-intercept form, if possible. 2(y – 1) + ((6x + 2)/3) = -8 (First, we need to get rid of the fraction, so multiply everything by 3.) 6(y – 1) + 6x + 2 = -24 (Now, distribute the 6 to the y and -1. Then, get y on a side by itself.) 6y – 6 + 6x + 2 = -24 6y = 6 – 6x – 2 – 24 6y = -6x – 20 y = -x – (10/3) (This is in proper linear form!) Now, find the equation of a line parallel to the line we just found that runs through the point (10, -13). y = -x – (10/3) (First, we need to substitute the values of x and y in the point into our equation. The “b” in slope- intercept form when we substitute these values will remain as “+ b”, we won’t use ( -10/3).) -13 = -10 + b -3 = b (Now, we have our b. Since the line is parallel, we’ll use the same slope as the original linear equation we found, which is -1.) y = -x – 3 (This is our parallel line!) EXAMPLE 5. Find the slope of a line that passes through the points (4, -8) and (9, 11). (To do this, we can use the formula (y – y )/(x – x ), so substitute the given values 2 1 2 1 into the formula.) (11 + 8)/(9 – 4) = (19/5) slope: (19/5) (This is our answer!) Now, find the slope-intercept form of this line, using this slope. (We just need to substitute the x and y values from one of the given points into the formula y = mx + b.) 11 = (19/5)(9) + b b = (-116/5) (Now, substitute this into the y = mx + b with the slope (m) and the b we just found.) y = (19/5)x – (116/5) (This is our answer!) EXAMPLE 6. o 2 Find (f g)(x), when f(x) = (x + 2)/5 and g(x) = x – 2. (To work an (f g)(x) function, just substitute the g(x) function everywhere you see an x in the f(x) function.) MATH121 Un niit Te esst 2 Leessssoonnss 3..2,, 3..3,, 3..4,, 4..1,, 4..5,, & 44..6 (f g)(x) = (x – 2 + 2)/5 (Now simplify.) o 2 (f g)(x) = (x /5) (This is our function!) Now, find (g f)(x) of the same functions. o 2 (g f)(x) = ((x+ 2)/5) – 2 (To simplify, our answer, we need to FOIL out the numerator of the fraction. Also, remember when we’re squaring a fraction, we square the numerator and the denominator.) o 2 (g f)(x) = (x + 4x + 4)/(25) – 2 (We can simplify this even more by making the -2 into a fraction with 25 as the denominator.) (g f)(x) = (x + 4x + 4)/(25) – (50/25) (Subtract the two fractions.) o 2 (g f)(x) = (x + 4x – 46)/25 (This is our answer!) EXAMLPE 7. Find the implied domain of this function. f(x) = (√(x + 1))/-4 (We need to look at the numerator and the denominator separately to determine at what points this function could be undefined. Remember, the implied domain is everywhere a function is defined. The numerator has a square root symbol. We should immediately recognize that what is under the sign is greater than or equal to zero. The denominator has no points of interest because it is a constant.) Domain: [ -1, ∞) (This is our answer!) EXAMPLE 8. 2 Find ((f(x + h) – f(x))/h), when f(x) = 4x + 5x – 5. (First, we need to substitute an (x + h) everywhere there’s an x in the f(x) function, then we need to subtract the original f(x) function from that. I usually leave the denominator (h) alone until a few steps later. Just don’t forget about it!) f(x + h) – f(x) = 4(x + h) + 5(x + h) – 5 – (4x + 5x – 5) 2 (Now, FOIL out the (x + h) then distribute the 4 to all the terms. Then distribute 5 to x and h. Don’t forget that when you subtract a bunch of terms, you’ll need to change their signs.) f(x + h) – f(x) = 4x + 8xh + 4h + 5x + 5h – 5 – 4x – 5x + 5 2 (Combine like terms.) 2 f(x + h) – f(x) = 8xh + 4h + 5h (Now is when we can bring the h back into play. Factor out an h from the function we just found.) ((f(x + h) – f(x))/h) = (h(8x + 4h + 5)/h) (The h in the numerator and in the denominator cancel each other out.) ((f(x + h) – f(x))/h) = 8x +4h +5 (This is our answer!) EXAMPLE 9. Find the inverse of the function P(x) = 3 √(x) – 5. 5 MATH121 Unniit Te estt 2 Leessssonns s 3..2,, 3.3,, 3..4,, 4..1,,4..5,, & 4..6 (The first thing we need to do is treat P(x) like y and then we need to switch the placement of the x and y variables in the problem.) P (x) = y = 3 √(x) – 5 x = 3 √(y) – 5 (Now, we need to isolate y. So first, add 5 to both sides.) x + 5 = 3 √(y) (Now divide both sides by 3.) (x + 5)/3 = √(y) (Raise both sides of the equation to the 5 power.) th ((x + 5)/3) = y (This is our answer!) EXAMPLE 10. Find the slope of the line that passes through the points (-7, 10) and (-5, 10). (Using the (y – 2 /x 1 x 2 form1la, substitute the values in.) (10 – 10)/(-5 + 7) = (0/2) (When the slope is a fraction with zero as the numerator, the slope is just zero!) Slope: 0 (This is our answer!) EXAMPLE 11. Find the implied domain of the function f(x) = (-13/(x – 12)). (We need to look at the numerator and the denominator separately to determine at what points this function could be undefined. Remember, the implied domain is everywhere a function is defined. We know that when a denominator is equal to zero, the function is undefined, so we know that x can equal to everything except 12 itself because 12 – 12 = 0.) Domain: (-∞, 12) U (12, ∞) EXAMPLE 12. Find (f g)(x), when f(x) = √(x + 1) and g(x) = x – 5. 2 (To work an (f g)(x) function, just substitute the g(x) function everywhere you see an x in the f(x) function.) (f g)(x) = √(x – 5 + 1) (Simplify.) (f g)(x) = √(x – 4) (This is our answer!) Now find (g f)(x) for the same functions. (g f)(x) = (√x + 1) – 5 2 (Simplify.) (g f)(x) = x + 1 – 5 (g f)(x) = x – 4 (This is our answer!) EXAMPLE 13. Put this equation into slope-intercept form. ((4x + 5)/2) – y = 3 – 2y (To get this in proper y = mx + b form, first you need to multiply everything on both sides of the equation by 2 to get rid of the fraction.) MATH121 Unni it Teesstt 2 Leessssoonnss 3..2,, 3..3,, 3..4,, 4..1,, 4..5,, & 4..6 4x + 5 – 2y = 6 – 4y (Get y on a side by itself.) 4x + 5 – 6 = -2y y = -2x + (1/2) Now, find the equation of a line perpendicular to the one we just found that runs through the point (11, 13). (First, we need to substitute the values of x and y in the point into our equation. The “b” in slope- intercept form when we substitute these values will remain as “+ b”, we won’t use (1/2). Since the new line is perpendicular to the original, we need to use the negative inverse of the slope we found for the original line.) 13 = (1/2)(11) + b (15/2) = b y = (1/2)x + (15/2) (This is our answer!) EXAMPLE 14. 3 1/3 Find the inverse of the function V(x) = (x + 3) + 3. (The first thing we need to do is treat V(x) like y and then we need to switch the placement of the x and y variables in the problem.) -1 3 1/3 V (x) = y = (x + 3) + 3 x = (y + 3) 1/3 + 3 (Now we need to isolate the y. First, subtract 3 from both sides.) x – 3 = (y + 3) 1/3 (We should realize that an exponent of 1/3 is the same thing as taking the cubed root. So, to get rid of the cubed root, raise both sides of the equation to the 3 power.) (x – 3) = y + 3 (Now, subtract 3 from both sides.) (x – 3) – 3 = y 3 (Take the cubed root of both sides of the equation to isolate y.) y = √((x – 3) – 3) (This is our answer!) EXAMPLE 15. 2 Find g(x + a) – g(x), when g(x) = -4x – 4x – 5. (First, we need to substitute an (x + a) everywhere there’s an x in the g(x) function, then we need to subtract the original g(x) function from that.) g(x + a) – g(x) = -4(x + a) – 4(x + a) – 5 – (-4x – 4x – 5) 2 (FOIL out the (x + a) and then distribute a -4 to all the terms in the quadratic. Then, distribute the -4 to x and a. Don’t forget, when subtracting multiple terms, you’ll need to change all of the signs.) 2 2 2 g(x + a) – g(x) = -4x – 8ax – 4a – 4x – 4a – 5 + 4x + 4x + 5 (Combine like terms.) g(x + a) – g(x) = -8ax – 4a – 4a 2 (This is our answer!) EXAMPLE 16. MATH121 Un niit Te esstt 2 Leessssoonnss 3..2,, 3..3,, 3..4,, 4..1,, 4..5,, & 44..6 Find (f + g)(x), when f(x) = x + 5 and g(x) = x 3/2 , then find the domain. (When adding two functions, it’s simple. Just stick a plus sign between them.) 3/2 (f + g)(x) = x + 5 + x (Now, to find the domain, we need to look for critical points of interest in the equation. Remember, x 3/2 is the same thing as √(x ). We 3 know that anything under a square root symbol must be greater than or equal to zero.) x ≥ 0 (This makes it easier to visualize the domain.) Domain: [0, ∞) Now, find (f/g)(x) for the same two functions. (f/g)(x) = (x + 5)/(x 3/2 ) = (x + 5)/(√(x )) 3 (We need to look at the numerator and the denominator separately to determine at what points this function could be undefined. Remember, the implied domain is everywhere a function is defined. We know that when a denominator is equal to zero, the functio n is undefined, so we know that x will be greater than zero.) x > 0 (This makes it easier to visualize the domain.) Domain: (0, ∞) EXAMPLE 17. Plot the line with this equation. -7y = -35 (Get y on a side by itself, since it is the only variable.) y = 5 (When y equals a constant, there will be a horizontal line that passes through (0, that constant).) EXAMPLE 18. Plot the line with this equation. 3y – 2x = 3y + 4 (Simplify so that one side of the equation equals zero.) 0 = 2x + 4 -4 = 2x -2 = x (When x equals a constant, there will be a vertical line that passes through (that constant, 0).) MATH121 Un niit Teesst 2 Leesssoonnss 3..2,,3..3,, 3..4,,4..1,, 4..5,,& 44.6 EXAMPLE 19. Plot the line with this equation. -2x = -2y (Divide both sides by -2.) x = y (When this occurs, we know there will be a positive slope because both variables are positive. Note: positive slopes start lower on the left, and go up towards the right. Negativ e slopes start higher on the left, and go down towards the right. Also, we know that it will pass through the point (0, 0).)

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