MATH 117 Pre-calculus Outline Part 1: Review of basic algebra and graphs, and Straight lines (constant slope) Part 2: Functions, quadratic functions (slope depends on x), and polynomials Part 3: Exponential functions (slope depends on y) and logarithms Part 4: Oscillating functions: sine, cosine, tangent; inverses; identities

Grading 4 tests (one over each part) each of which counts as one grade Computer homework (8 assignments) the total of which counts as one grade Regular written homework (9 assignments) the total of which counts as one grade WebCT tests (7 assignments) which together count as one grade. Final Exam which counts as three grades. Final Grade is determined by averaging the 10 grades.

Important Note You must earn at least a “C” or better grade in Math 117 to be eligible to move on to Math 131 Calculus I at CBU. Also, most institutions do not transfer courses with grades less than “C”.

Usual notation We will be using a lot of symbols in this course. Sometimes we will use symbols based on the first letter of a word, such as r for radius. Otherwise, we will tend to use n and m for integers, x for the independent variable, y for the dependent variable, and the first letters of the alphabet for constants. Also we will assume that the SQRT notation always gives a positive. If we need to show that the square root of 4 is both +2 and –2, we will write it as ±SQRT(4) = ±√[4].

Basic Algebra Real numbers (on the number line) –Rational numbers: r = n/m (where m is not 0, and n and m are both integers) –Irrational numbers: other numbers (for example, π and e) –Counter example: Square root of a negative number is not a real number

Basic Algebra Decimal equivalence of fractions: –Examples: 4/5 = 0.8 ; 4⅓ = … –Going backwards: d = … now note that 1000*d = … now subtract: 1000*d – d = 125 so 999*d = 125, or d = 125/999 Scientific notation: –Examples: 1,500,000 = 1.5 x 10 6 ; = 2.5 x Calculators and approximations

Basic Algebra Interval notation: –[3,5] means the interval that DOES include the end points of 3 and 5. –(3,5) means the interval that DOES NOT include the end points of 3 and 5. –Example: Newton’s Law of Gravity says F = G*M*m/r 2 which is good for (0,∞) since r is the distance between two masses, M and m, and r can’t be zero - it can only approach zero.

Basic Algebra Absolute value: |x| If x ≥ 0, then |x| = x; if x < 0, then |x| = -x. Example of use: “distance” |x - a| = d given a (starting location) & d (distance, must be > 0), solve for the two positions, x: x = a + d, and x = a – d. Example: |x - (-4)| = 7: x = = 3, and x = = -11. Check: |3-(-4)| = 7 and |-11-(-4)| = 7.

Basic Algebra A direct relation is one where as one variable increases, a related variable also increases. An example would be y = 5x. A physical example is the pressure of the water increases directly as the depth of the water increases: P = ρgh (where P is the pressure and h is the depth). An inverse relation is one where as one variable increases, a related variable decreases. An example would be y = 5/x. A physical example is Newton’s Law of Gravity: F = GMm/r 2, as the distance from the center of the earth, r, increases, the Force, F, decreases as 1/r 2. The force is inversely proportional to the square of the radius.

Basic Algebra Solving linear equations: – ax + b = cisolate the unknown (x) divide every term by a: x + b/a = c/a; now add –b/a to both sides: x = c/a – b/a. –example: 5*x – 8 = -4 divide thru by 5: x – 8/5 = -4/5 add a negative –8/5 to both sides: x = -4/5 - -8/5 x = -4/5 + 8/5 = 4/5. Check: 5(4/5) – 8 = -4 becomes 4 – 8 = -4.

Basic Algebra Solving quadratic equations: – ax 2 + bx + c = 0 write as: ax 2 + bx = -c divide thru by a: x 2 + (b/a)x = -c/a complete the square using fact that (x+f) 2 = x 2 + 2fx + f 2, so need to add (f) 2 to each side where here 2f = b/a: x 2 + (b/a)x + (b/2a) 2 = -c/a + (b/2a) 2 so that we have [x+(b/2a)] 2 = -c/a + b 2 /4a 2 taking the square root and re-arranging gives: x = [-(b/2a) ± SQRT{(b 2 /4a 2 ) – (4ac/4a 2 )} factoring out the 4a 2 from the denominator of the square root gives the standard form: x = [-b ± SQRT{b 2 – 4ac}] / 2a.

Basic Algebra Besides linear and quadratic equations, we can keep on upping the power of x. Equations of the form: ax n + bx n-1 + … + ux 2 + vx + w = 0 are called polynomial equations of degree n. Example: 17x 4 + 3x 3 – 8x 2 – 9x = 11 is a polynomial of degree 4. Usually these are solved graphically or numerically, although sometimes we can use tricks to solve some of the special cases. One special case is an equation of the form: ax 4 + bx 2 + c = 0. Here we simply use u=x 2, to get au 2 + bu + c = 0; solve for u, and then solve for x.

Computer Homework You should be able to do the first computer homework program on Conversion Factors (Volume #1, #1). Note that when you work with applications, the values usually have units that go with the values. This means that instead of using abstract variables like x and y, you will be using physical or financial variables like pressure, temperature, time, or money that have units.

Graphs The number line was a nice “picture” of what we will call a 1-dimensional system. What about a plane? Can we identify a position in a plane? What about your address? Where is Memphis on a world map? Cartesian (or rectangular) coordinates are one way: (x,y)

Graphs Points in a plane can be specified by their coordinates: (x,y). For instance, a certain point, A, will be specified by (x A, y A ), and another point, B, will be specified by (x B, y B ) x y

Graphs If point A is specified below, what are the coordinates of point A? If point B has coordinates of (-10,-15 ), where would point B be located? A y x

Graphs The horizontal is usually specified as “x”, and the vertical as “y”. This makes the coordinates of point A (20,10) and point B (-10, -15) A B y x

Graphs Can we define a “distance”, d, between points A and B? How do we find (calculate) that distance? A B y x d

Graphs By using the Pythagorean Theorem, we have: d = SQRT[(x A -x B ) 2 + (y A -y B ) 2 ] Does it matter if we use (x A -x B ) or (x B -x A ) ? A B y x d

Graphs For our example with A at (20,10) and B at (-10,-15), we have d = SQRT[(20-[-10]) 2 + (10-[-15]) 2 ] = SQRT ( ) = A B y x d

Graphs Besides graphing points, we can graph equations that relate y to x, since an equation is a relation between values of y and values of x. This graph, then, is a picture of the solutions of the equation y x

Graphs The definition of a circle is that the points are all equidistant from the center. We just defined distance on a previous slide, so if we start from (x c,y c ) as the center point, then the equation should be: d = SQRT[(x-x c ) 2 + (y-y c ) 2 ], or (x-x c ) 2 + (y-y c ) 2 = d 2

Graphs Example: Let’s choose the center point of the circle to be (-10, 5), and the distance to be 15. Our equation is then 15 = SQRT[(x-(-10)) 2 + (y-5) 2 ], or (x+10) 2 + (y-5) 2 = y x d=15

Graphs Let’s look at the graph of ax + by = c, and see if we can notice anything special. To be definite, let’s choose a = 4, b = -7, and c = -8. The equation in this case is 4x – 7y = -8.

Graphs Example: 4x – 7y = -8 is an equation. If we choose y=0, then the solution for x is x=-2. If we choose y=5, then x=6.75. If we choose y=15, then x= It looks like the points fall on a straight line! y x

Graphs Note that in both the graphs for the circle and the straight line, we could see where the circle or line cut across the x axis. This (or these) is (are) the value (s) of x that satisfies the equation when y=0. Thus, if we want to solve an equation for x when y=0, we can graph the equation and then look to see what the value (s) of x is (are) when the curve/line crosses the x axis!

Graphs Example: From a previous slide, we plotted a circle with the equation: (x+10) 2 + (y-5) 2 = If we set y=0, we have the equation (x+10) 2 + (-5) 2 = By looking at the graph, we can see that there are two solutions, one near x=4 and the other near x= y x d=15

Graphs Example: From a previous slide, we plotted a line with the equation: 4x – 7y = -8. If we set y=0, we have the equation 4x – 0 = -8. By looking at the graph, we can see that there is only one solution, near x= y x

Graphs Notice that the circle equation, (x-x c ) 2 + (y-y c ) 2 = d 2, was a second degree equation and had the possibility of having two solutions. Notice that the straight line equation, 4x – 7y = -8, was a first degree equation and had the possibility of having one solution. Is the generalization that an n-th degree equation has the possibility of having n solutions true?

Graphs If we have an equation of the form: ax + by = c, can you recognize this as a straight line? If we have an equation of the form: x 2 + bx + y 2 + dy = f, can you recognize this as being any special type of line? Could you put it in “standard form” for that special type? [Hint: complete the square for both x and y.]

Graphs Example: Given x 2 + 3x + y 2 - 4y = 11, put this in the form of a circle. First we complete the square for the x: x 2 + 3x + (3/2) 2 + y 2 - 4y = 11 + (3/2) 2, or (x+1.5) 2 + y 2 - 4y = Now we complete the square for the y: (x+1.5) 2 + y 2 - 4y + (4/2) 2 = (4/2) 2, or (x+1.5) 2 + (y-2) 2 = where we can identify x c = -1.5, y c = 2, and d = SQRT(17.25).

Graphs What would happen if we multiplied the x 2 term by a constant other than 1? What would happen if we multiplied the y 2 term by a constant other than 1? What would happen if either of those constants were negative? What would happen if both of those constants were negative?

Graphs Example: What about the graph of absolute value: |x - a| = d ? Can you “picture” what this will look like? What do the constants a and d do for the graph?

Graphs x Given: y = |x - (-4)| (here d = y, and a = -4) when x = 0, y = 4 when x = 2, y = 6 when x = 4, y = 8 when x = -2, y = 2 when x = -4, y = 0 when x = -6, y = 2 when x = -8, y = 4 y 8-8 8

Regular Homework set #1 (continued on next page) 1.Find the fractional expression (s/r) for the rational number … (where the sequence continues to repeat) For problems 2-7, solve all for x, and then check your answer: 2. |x + 6| = x + 4 = x 2 + 4x + 7 = 0 5. You choose non-zero values for a, b, and c; and solve ax 2 + bx + c = 0 for x, and check your answer.

Regular Homework Set #1 (Continued from previous page) 6. -4x 2 + 7x – 3 = 8 7. [3/(x+5)] = [4/(x-7)] Hint: get a common denominator, and as long as the denominator is not zero, you only have to make the numerator zero. 8.Given the equation x 2 + 9x + y 2 – 4y = 20, find the radius and the center of the circle. 9.Given the center of a circle is at (2, -7), and the circle has a radius of 12, determine the equation relating x and y. 10.Where (what values of x) does this circle of #9 cross the x axis?

Computer Homework You should be able to do the second computer homework program Relations (Volume 0, #1). Note that when you graph data in applications, the values usually have units that go with the values. This means that instead of using abstract variables like x and y, you will be using physical or financial variables like pressure, temperature, time, or money that have units. That means that the axes for graphs will also have units with them.

Graphing Calculators We’ll now spend time with the graphing calculators. Diamond / F1 (Y=) Use up and down areas and CLEAR button to clear out any old functions Type in: 2x+3 then press diamond / F3 (graph) Diamond / F2 (Window) shows min & max x and y and scales used. Adjust these to see area of interest. Note that x is about twice as long as y is high on display.

Graphing Calculators In looking at graphs of relations, we can look for different kinds of symmetries (samenesses). Does the graph look the same on the left side of the y axis as it does on the right side (mirror reflection); if so (x,y) = (-x,y). Does the graph look the same above the horizontal (x) axis as it does below (mirror reflection); if so (x,y) = (x,-y). Does the graph look the same through the origin; if so (x,y) = (-x,-y)

Regular Homework set #2 1.Graph the equation: y = x 3 – 5x +2x +4, on your calculator, find the local max and min by reading your graph, and find by reading the graph the values of x where the curve crosses the x axis (where y=0). 2.Graph the two equations: y = 2x 4 – 3x and y = x 3 – 5x + 2x + 4. Make a sketch of the plot from the calculator. By reading your graph, find all approximate values of x where the two curves intersect.