x coordinates y coordinates Compare all the x coordinates, repeats. The set is not a function, just a relation. Compare all the x coordinates, no repeats. The set is a function.

Compare all the x coordinates in the domain, only one corresponding arrow on each x coordinate. The set is a function. Compare all the x coordinates in the domain, 8 has two corresponding arrows. Repeats The set is not a function, just a relation.

When determining if a graph is a function, we will use the Vertical Line Test. Use your pencil as a Vertical Line and place it at the left side of the graph. Slide the pencil to the right and see if it touches the graph ONLY ONCE. If it does it is a FUNCTION. FUNCTION. Use your pencil as a Vertical Line and place it at the left side of the graph. The Vertical Line crosses the graph in 2 or more locations, therefore this graph is just a RELATION.

How to write equations as a function? Solve for y! Implicit form Mr. Fitz’s Rule! – 2x = – 2x 1. y can not be raised to an even power! y even +…. 2. y can not be in absolute value bars! | y | + …. Explicit form y is to an odd power … solve for y. – x3 = – x3 y is to an even power … NOT A FUNCTION! Replace y with g(x).

y = 3(4) + 7 y = 12 + 7 y = 19 The work is the same! f(4) = 3(4) + 7 y coordinates input output y = f(x) Dependent Variable Independent Variable y = 3(4) + 7 y = 12 + 7 y = 19 The work is the same! f(4) = 3(4) + 7 f(4) = 12 + 7 f(4) = 19

Put ( )’s around every x. Substitute -6 for every x. 10 Simplify by Order of Operations. FOIL and distribute Combine Like Terms, CLT.

Remember h(x) = y ??? h(x) = y h(x) = y h(x) = y h(3) = 2 h(2) = 1 h(0) is not possible! Zero is not in the Domain. Undefined = y = y = y (3, 5) Find the point when x = 3 Find the point when x = -2 Find the point when x = 0 (-2, 1) j(3) = 5 j(-2) = 1 j(0) = -1 3 -2 (0, -1)

[3, 6] interval notation = y x = 2 x = -4, -2 & 1 3 < x < 6 Every x coordinate from 3 to 6 = y (3, 5) (6, 5) 5 Find the point when y = 3 Find the point when y = 1 3 (2, 3) (-4, 1) x = 2 x = -4, -2 & 1 1 (-2, 1) (1, 1) -3 (-?, -3) (?, 3) Find the point when y = 5 Find the point when y = -3 3 < x < 6 j(x) = -3 is not possible! -3 is not in the Range. Undefined [3, 6] interval notation

Domain Find the smallest x coordinate to the largest x coordinate. 7 5 Domain: -7 < x < 6 or [-7, 6] 3 -7 Range Find the smallest y coordinate to the largest y coordinate. 6 The first set of y coordinates are -4 < y < 3 or (-4, 3). Notice that we started and ended at open circles. -4 The second set of y coordinates are 5 < y < 7 or [5, 7] Open circles mean that the point doesn’t exist and the closed circle means that the point is there. x = -3 at this location…as long as we can touch the graph the x coordinates are there and continuous. Range: -4 < y < 3 or 5 < y < 7 (-4, 3) U [5, 7]

Domain Find the smallest x coordinate to the largest x coordinate. Domain: x > -4 or [-4, oo) Range Find the smallest y coordinate to the largest y coordinate. -4 Range: y > -7 or [-7, oo) -7

4 1 -8 8 -1 Domain: {-1, 1, 2, 3, 4, 5} Range: {1, 2, 3, 4, 7, 8} Find the smallest x coordinate to the largest x coordinate. 4 Domain: -8 < x < 8 or [-8, 8] 1 Range Find the smallest y coordinate to the largest y coordinate. -8 8 -1 The y coordinates are not connected or consistent, therefore we list them separately. Range: {-1, 1, 4} When given the function in set notation, list the x and y coordinates separately. Domain: {-1, 1, 2, 3, 4, 5} Range: {1, 2, 3, 4, 7, 8}

Find the domain of the functions. When finding the domain of functions in equation form we will ask ourselves the following questions…. Will the function work when the x is a negative?, …. a zero?, … a positive? If the answers are 3 yes’s, then the domain is all real numbers. If there is a no, then there is a domain restriction we need to find. Can I multiply 4 by a negative?, a zero?, a positive? … and then add 2 to the product? ALL Yes! Domain is ALL REAL NUMBERS Can I square a negative?, a zero?, a positive? … and then add 2 to the value? ALL Yes! Domain is ALL REAL NUMBERS If I square a negative?, a zero?, a positive? … I should be able to raise them to any positive power! ALL Yes! Domain is ALL REAL NUMBERS

Find the domain of the functions. Adding and subtracting always is a Yes…Can I divide by a negative?, a zero?, a positive? NO! Can’t divide by ZERO! Set the denominator equal to zero and solve for x to find the restriction. Domain is ALL REAL NUMBERS, except 1 Can I take the absolute value of a negative?, a zero?, a positive? … and then subtract 9 to the value? ALL Yes! Domain is ALL REAL NUMBERS We have a fraction again, set the bottom equal to zero and solve for x. Domain is ALL REAL NUMBERS, except for -3 and 3.

We have a fraction again, set the bottom equal to zero and solve for x by factoring. Domain is ALL REAL NUMBERS, except for -8 and 2. We have a fraction again, set the bottom equal to zero and solve for x by factoring. Domain is ALL REAL NUMBERS, except for -3, 0 and 4.

Domain Restrictions 5 < -5, FALSE -5 < 5 < 3 , FALSE 5 > 3, TRUE x = 5, test it in the domain restrictions to see which one is true! Substitute the 5 into that function. x = 3, and 3 > 3. Substitute 3 into the third function. x = -7, and -7 < -5. Substitute -7 into the first function. x = -5, and -5 < -5 < 3. Substitute -5 into the second function.

Cubic Func.

(0, 6) down 5 right 2 starting point y-int = (0, 6) down 5 directions rise run m = slope = down 5 b = y-int = (0, b) right 2 starting point y-int = (0, 6) down 5 directions -5 2 m = right 2

(-3, 4) starting point (-3, 4) directions 1 3 m = Or in reverse right 3 up 1 right 3 up 1 right 3 point = (x1, y1) left 3 up 1 rise run down 1 m = slope = (-3, 4) starting point (-3, 4) directions 1 3 m = Or in reverse

(0, y) ( x, 0) A, B, and C are integers. To graph find x and y intercepts ??? To find the y intercept the x coordinate is zero! (0, y) To find the x intercept the y coordinate is zero! ( x, 0) Doesn’t fit, but that is ok…we can use the slope!

Notice that there is no y variable in the equation Notice that there is no y variable in the equation. This means we can’t cross the y axis! Must be a VERTICAL LINE at x = 6 rise m = slope = = undefined Notice that there is no x variable in the equation. This means we can’t cross the x axis! Must be a HORIZONTAL LINE at y = - 4 run m = slope = = 0

( x, 0) To graph find x and y intercepts. We can see that 3 will divide into -9 evenly, but 5 won’t. So we should find the x intercept and the slope to graph this line. To find the x intercept the y coordinate is zero! ( x, 0) Find the slope!

They are all good! Here is the work for all 3 forms. Write the equation of a line that contains the points (3, 8) and (5, -1). In all 3 forms, the slope is present. Find the slope between the points. 2nd y coord. minus 1st y coord. 2nd x coord. minus 1st x coord. Which of the 3 forms of the equations should we use to finish the problem? They are all good! Here is the work for all 3 forms. Using point (3, 8). Slope intercept Solve for b by plugging in one of the given points for x and y. I will use (3, 8) Point Slope form The Frac command is in the MATH button, #1 DONE! Standard form A = 9 B = 2 Using point (3, 8).

Which equation should we use? Write the equation of a line that is parallel to 4x + 7y = 11 and contains the point (-2, 3). Since there is an equation given in standard form, stay with it! Which equation should we use? They never cross each other because the SLOPES are the same! What do we know about parallel lines? Our new equation has to have the same A and B because these numbers create the slope! We need a new C! Given. NEW. Plug in (-2, 3) for x and y to solve for C! MyMathLab will probably want the equation written in y = mx + b. Solve for y!  

What do we know about perpendicular lines? Write the equation of a line that is perpendicular to 4x + 7y = 11 and contains the point (-2, 3). What do we know about perpendicular lines? The slopes are opposite reciprocals! Our new equation has to have A and B switched and a change in the sign. Given. NEW. Proof Plug in (-2, 3) for x and y to solve for C!        

Remember…functions are equal to y. y = C(m). Use y = mx + b. Yellow TAXI Cab Co. charges a $10 pick-up fee and charges $1.25 for each mile. Write a cost function, C(m) that is dependent on the miles, m, driven. Remember…functions are equal to y. y = C(m). Use y = mx + b. The slope is the same as the rate! The y intercept (b) is the starting point or initial cost. The $10 pick-up fee is a one time charge or initial cost. b = 10 The $1.25 for each mile is a rate. m = 1.25 Replace y with C(m) and x with m.

Estimate the life expectancy in the year 2009. In the year 2000, the life expectancy of females was 83.5. In 2004, it was 86.5. Write a linear function E(t) where t is the number of years after 2000 and E(t) is the life expectancy in t years. Estimate the life expectancy in the year 2009. Estimate when the life expectancy will be 94. Looks difficult only because of all the words! Understand the data given to write the equation of a line! This looks like points (x, y) = (t, E(t)) Year # of years after 2000 (t) Age E(t) 2000 0 83.5 2004 4 86.5 (0, 83.5) (4, 86.5) We are back to the first problem we did for writing the equation of a line. Use y = mx + b because we are working with functions and (0, 83.5) is the y intercept….b is 83.5. Find the slope between the points. Estimate the life expectancy in the year 2009. Estimate when the life expectancy will be 94.     14 years past the year 2000, 2014.

Understand the data given to write the equation of a line! In the year 2003, a certain college had 3450 students. In the year 2008, the college had 4100 students. Write a linear function P(t) where t is the number of years after 2000 and P(t) is the population of the college. Estimate the population in the year 2012. Estimate the year when the population will reach 5400. Understand the data given to write the equation of a line! Year # of years after 2000 (t) Students P(t) 2003 3 3450 2008 8 4100 Points (x, y) = (t, P(t)) (3, 3450) (8, 4100) Use y = mx + b because we are working with functions, but this time we will have to solve for b. Find the slope between the points. Plug in a point, (8, 4100). Estimate the population in the year 2012. Estimate the year when the population will reach 5400. 18 years past the year 2000 is the year 2018.

= f(2) + g(2) = f(5) – g(5) = f(-2) g(-2) = [(2)2 + 9] + [3(2) + 5] QUOTIENT: = f(2) + g(2) = f(5) – g(5) = f(-2) g(-2) = [(2)2 + 9] + [3(2) + 5] = [(5)2 + 9] – [3(5) + 5] = [(-2)2 + 9] [3(-2) + 5] = [4 + 9] + [6 + 5] = [25 + 9] – [15 + 5] = [4 + 9] [-6 + 5] = 13 + 11 = 34 – 20 = 13(-1)

[(1)2 + 9] [3(1) + 5] [x2 + 9] [3x + 5] [3x + 5] [x2 + 9] f(1) g(1) = = f(x) + g(x) = g(x) – f(x) = [x2 + 9] + [3x + 5] = [3x + 5] – [x2 + 9] [(1)2 + 9] [3(1) + 5] 10 8 = Distribute the minus sign. = = 3x + 5 – x2 – 9 f(x) g(x) g(x) f(x) = f(x) g(x) = = = [x2 + 9] [3x + 5] [x2 + 9] [3x + 5] [3x + 5] [x2 + 9] = = F O I L Domain Restriction Not possible! No Domain Restriction

Multiple Domain Restrictions! AARRRGGGG! WAIT! We don’t divide by fractions, we… multiply by the reciprocal! WAIT! There is a new factor in the denominator!

[-9, 4] 4 -9 [-14, 4] [-9, 11] -9 < x < 4 -14 4 The Domain when adding, subtracting, or multiplying is where the two graphs OVERLAP! [-9, 4] -9 < x < 4 If you were to write the four endpoints in numerical order, it will be the two middle numbers. -14, -9, 4, 11

[-9, 4] with no restrictions. S(x) This means when y = 0, or the x intercepts on S(x). In the OVERLAPPING Domain. (8, 0) isn’t in the Overlapping Domain, so throw it out (-2, 0) (8, 0) [-9, 4]; [-9, -2) U (-2, 4] This means when y = 0, or the x intercepts on R(x). In the OVERLAPPING Domain, but R(x) doesn’t cross the x axis. [-9, 4] with no restrictions.

(0, 5.6) (-4, 5) (-7, 4) (0, 2) -4 -4 -7 -7 (-4, -2) (-7, -4) = R(0) + S(0) = 5.6 + 2 = R(-7) – S(-7) = R(-4) S(-4) = 7.6 = 4 – (-4) = 5 (-2) = 8 = -10

This means to get the indicated variable by itself. Find the indicated variable. Find the indicated variable. Remove all fractions. Remove all fractions. Multiply by (s + v) to both sides.             Distribute f and get s terms to one side. Isolate the h. Divide by V2 Factor s as the GCF and divide by (g – f). Isolate the h. Subtract R.

Find the indicated variable. Remove all fractions. Multiply all fractions by LCD = abc Cancel all denominators and get the a terms to one side. Factor a as the GCF and divide by (b – c).

Direct Variation Inverse Variation Joint Variation Variations Have a variation konstant, k, in the formulas that we will need to find. Direct Variation Inverse Variation Joint Variation k is multiplied by 2 or more variables. k is multiplied by x. k is divided by x. “y varies directly as x.” “y varies inversely as x.” “y varies jointly as x and w.” Variations will have the following steps to solve. 1. Write the variation equation. Will be in the first sentence or statement. 2. Solve for k and rewrite the formula. Information will be in second sentence or statement. 3. Solve for the indicated variable. Information will be in third sentence or statement. Y varies directly as the square of x. If y = 9 when x = 6, then what is the value of y when x = 8. 3. y = 0.25 (8)2 y = 0.25(64) y = 16 2. 9 = k (6)2 9 = 36 k 0.25 = k y = 0.25 x2 1. y = k x2

M varies inversely as the square root of t M varies inversely as the square root of t. If M = 3 when t = 4, then what is the value of M when t = 81. B varies jointly as c and d cubed. If B = 120 when c = 3 and d = 2, then what is the value of B when c = 25 and d = 0.2.

Y varies directly as the square of x and inversely as w Y varies directly as the square of x and inversely as w. If Y = 8 when x = 2 and w = 5, then what is the value of Y when x = 6 and w = 30.