Greg Kaulius, Zach Skank, and Richie Donahue

Points and lines

X-axis 2 nd Quadrant 4 th Quadrant3 rd Quadrant 1 st Quadrant S(-6,4) Y-axis X-axis S point X-coordinate is -6 y-coordinate is 4

Ex How to Sketch 2x+5y=20 1 st Let x=0, solve 2(0)+5y=20 5y=20 y=4 (0,4) Let y=o, solve 2x+5(0)=20 2x=20 x=10 (10,0) 2 nd Plot the points (0,4) and (10,0) Draw a straight line between and Pick a point that lies on the line. 3 rd Check points (5,2) Plug numbers into problem 2(5)+5(2)=20 √ (0,4) (10,0) (5,2)

1 st Get X to Equal (4)2x+3y=15 8x+12y=60 (2)4x-9y=3 8x-18y=6 2 nd Subtract 2 equations: 8x+12y=60 8x-18y=6 30y=54 y=9/5 3 rd Plug 9/5 in for Y 2x+3(9/5)=15 x=24/5 Point of intersection is (9/5,24/5) Ex Find where the 2 lines intersect 2x+3y=15 4x-9y=3

AB Ex Given C(1,0) D(7,8) 1 st Plug in the numbers to Distance equation =√(7-(1))²+(8-(0)) ² = √36+64 Distance=10 Ex Given C(3,3) D(15,12) 1 st Plug in numbers to Distance equation =√(15-(3))²+(12-(3)) ² =√ Distance=15

Midpoint Formula Ex Given C(3,3) D(15,12) 1st Plug in numbers to Midpoint equation 2 nd Solve=(3+15,3+12) ( 2 2 ) Midpoint= (9,9) Ex Given C(1,0) D(7,8) 1st Plug in the numbers to Midpoint equation 2 nd solve=(1+7,0+8) ( 2 2 ) Midpoint=(4,4)

Slopes of Lines

Slope intercept Form Y=mx+b M=rise=y 2 -y 1 run x 2 -x 1 slope Ex (5,3) (2,0) M=3-0 =3 = Ex (0,4) (6,0) M=3-0=

 If a line is horizontal it was a slope of 0.  If a line is vertical it doesn’t have a slope.

 2 nonvertical lines are parallel if and only if they have the same slope Y=2x+6 Y=2x+2 Y=2x-4

 2 lines are perpendicular if and only if their slopes are negative reciprocals of each other  M1= -1, or m 1 xm 2 =-1 m 2 Ex y=x+2 y=-x+3 These numbers are negative reciprocals

Finding Equations of Lines

Linear Equation Forms: The General form Ax+By=C The slope- intercept form y= mx+k Line has slope m and y-int k The point- slope y-y 1 =m x-x 1 Line has slope m and contains (x 1,y 1 ) The Intercept form x + y = 1 a b Line has x-int a and y-int b.

Example Using intercept form Find an equation of the line with x-int 20 and y-int 8 Intercept form: x + y = 1 a b 1 st since you got the x and y-int use the intercept form plug in the x and y int for a and b in Equation x + y = nd Answer in general form x+4y=20

Example Using Point-Slope form The line through (-1,4) and (5,8) Point slope form: y-y1=m x-x1 1 st Find the slope M= 4-(8) -1-5 =2/3  2 nd Using the point slope form plug the slope in for M and either (- 1,4) or (5,8) in for x 1 and y 1 When using (-1,4):When using (5,8): y – 4 = 2 y – 8 = 2 x-(-1) 3 x – 5 3 In General Form=2x-3y=-14

Example Using Slope-Intercept form  The line with y-int 1.8 and parallel to.3x-1.2y=6.4  Slope-intercept form: y= mx+k  1 st Write the equation.3x-1.2y=6.4 in slope-intercept form to get y=.25x – 5.33  2 nd Use the slope-int form y=.25x + 3  3 rd Answer in general form.25x – y = -1.8

 Function describes a dependent relationship between quantities  Linear Functions have the form f(x)= mx + k  This is read “f” of “x” f(x) = 3x – 5 f(2) = 1 f(5/3) = 0 so we say that 5/3 is a zero of the function f

Barbara's boat costs $700 a month for payments and insurance. Gas and maintenance cost $0.25 a mile. a. Express total monthly cost as a function of miles b. What is the slope of the graph of the cost function A. C(m) =.25m B. Slope =.25 = 1/4

Section 1-5 Complex Numbers

Real Numbers  Real numbers are number which can be found on a continuous number line Complex Numbers  Also known as imaginary numbers  Ex. i = √ -1 and √-a = i √a

Practice with complex numbers 1. √7 ∙ √2 2. √ -7 √14 i√7 3. √-4 + √-16 + √-14.= √-25 i√4 + i√16 + i√1 √-50 2i + 4i + I = i√25 7i i√50 i√(1/2) = i√2 2

Solving Quadratic Equations

Quadratic Equation- It’s any equation that can be written in the form where a ≠ 0. A root of a quadratic equation is a value of the variable that satisfies the equation.

You can solve these in 3 ways- 1.Factoring- whenever the product of two factors is zero, at least one of the factors must be zero. 2. Completing the square- The method of transforming a quadratic equation so that one side is a perfect square trinomial 3. Quadratic formula- derived by completing the square.

Examples  Solve y²-8y=2  y²-8y-2=0  Use quadratic equation  -b+- √b²-4ac 2c  8+- √64-4(1)(-2) 2  8+- √72 2  8+-6 √2 2  4+-3 √2  Solve (4x-1) ²=-4  16x²-8x+5=0  Use quadratic equation -b+- √b²- 4ac 2c 8+- √64-4(16)(5) √ i i 8

Quadratic Functions and their Graphs

Quadratic function- is the set of points (x, y) that satisfy the equation. axis of symmetry- when you fold the graph along this axis, the two halves of the graph coincide. The vertex of the parabola is the point where the axis of symmetry intersects the parabola.

Find the intercepts, axis of symmetry, and vertex of the parabola y = (x + 4)(2x – 3). Axis of Symmetry : Equation=-b 2a = -5 4 Vertex x : -5 4 To get y plug -5/4 in for x y=2(-5/4) ²+5(-5/4)-12 =-9/8 X intercept: plug 0 in for x y=2(0)²+5(0)-12 =12 Y intercept: plug 0 in for y 0=2x²+5x-12 =-4 3/2

 Curve fitting-it is the process in which you find many kinds of curves to pass through data points.  One possible curve is the parabola with equation C(x)=ax2+bx+c  To find the values of a, b, and c, substitute the data from the graph into this equation.

 Using the given model, determine the number of calories burned per gram hour by a parakeet flying level at 26mi/h. Compare your answer with the actual laboratory result of calories per gram hour. 1 st substitute the speed of 26 mi/h. into the quadratic equation. c(x)=0.678x ² x c(26)=.678(26) ² (26) c(x)= The energy expenditure of the parakeet flying at 26mi/h is about 124 calories per gram hour. Since this result is approximately within one unit of the actual energy expenditure of calories per gram hour, we can assume that our model is probably a good one.