MAT092::Lecture Note::Week 06

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Overview

Linear functions.

Assessments:

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Linear Functions

M-W.com defines linear as follows.
   "of, relating to, resembling, or having a graph that 
    is a line and especially a straight line"

   [...and...]

   "having or being a response or output that is directly 
    proportional to the input"
Observe that the word linear contains the word line.

A linear function is a function is a "first degree polynomial function of one variable. These functions are called 'linear' because they are precisely the functions whose graph in the Cartesian coordinate plane is a straight line."

The following are equations for a line.
   slope-intercept form:  y = mx + b
   ... used when slope and y-intercept are known

   point-slope form:  y - y₁ = m(x - x₁)
   ... where (x₁, y₁) is a point on the line having slope m

   standard form:  Ax + By = C
Slope is another way of saying "rate of change" and linear functions have a constant rate of change.

Slope is often described as the ratio "rise over run."

Linear functions with a positive constant rate of change are increasing functions. Linear functions with a negative constant rate of change are decreasing functions.

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Slope (Linear Functions)

The slope of a line is a measure of its average rate of change (steepness?). The slope also indicates if the line is increasing (uphill?) or decreasing (downhill?). In linear equations, slope is represented by the letter m.

The following is how to find the slope between two unique points on a line.
   point 1:  (x₁, y₁)
   point 2:  (x₂, y₂) 

   slope = m = (y₂ - y₁) / (x₂ - x₂)

   note:  x₁ ≠ x₂
Slope represents the ratio of change-in-output over change-in-input.
   m = Δy / Δx    (change in output / change in input)

   Δ is the "delta" character
   Δy is read "change of" (Δy is read "change of" y)
   Δy = y₂ - y₁
Slope is often described as the ratio "rise over run."
        rise
   m = ------
        run
Some slope notes.
   horizonal lines have slope 0   (y = f(x) = k, where k is a consant)
   vertical lines have undefined slope   (x = k, where k is a consant)
   parallel lines have equal slopes     (m₁ = m₂)
   perpendicular lines have negative inverse slopes  (m₂ = -1/m₁)
The following is a verbose form the slope-intercept equation.
   output = slope * input + initial_value_if_any

   ...or...

   output = average_rate_of_change * input + initial_value_if_any

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Intercepts (Linear Functions)

Intercepts are points where lines intersect. In many cases, we need to know where lines intersect the horizonal- and vertical-axis (often called the x-intercept and y-intercept, respectively).

The vertical-intercept for a function is found by evaluating a function when its input is zero.

The horizontal-intercept for a function is found by setting its output to zero and finding what input value results in that output.
f(x) = 5(x) + 3 vertical-intercept (y-intercept)... set the input to 0 and evaluate: f(0) = 5(0) + 3 = 3 vertical-intercept is (0, 3) horizontal-intercept (x-intercept)... set the output to 0 and solve: 0 = 5(x) + 3 -3 = 5(x) -3/5 = x horizontal-intercept is (-3/5, 0)

Generalizations.
given f(x) = m(x) + b vertical-intercept: (0, b) [notice input is 0] horizontal-intercept: (-b/m, 0) [notice output is 0]

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Exercise: Find Equation of a Line Given Two Points

Memorize the following

   slope formula:  m = (y₂ - y₁) / (x₂ - x₁)
   slope-intercept form:   y = mx + b
   point-slope form: y - y₁ = m(x - x₁)

[exercise] Given the two points (1, 5) and (3, 15), find the equation of the line.

step 1: calculate the slope m (15 - 5) / (3 - 1) = 10 / 2 = 5 step 2: substitute one of the points and the slope into the point-slope form equation for a line y - 5 = 5(x - 1) # point (1, 5) selected; m=5 y₁=5, x₁=1 step 3: solve for y y - 5 = 5x - 5 # distributive property on right-side y = 5x - 5 + 5 # add 5 to both sides y = 5x + 0 # this is slope-intercept form

Since the 'b' value is zero we know that this line passes through point (0, 0). Recall, the 'b' value is the y-intercept (vertical-intercept) and it is obtained by setting the input to zero.

Solving this problem using the TI-83 calcuator.

press ON press STAT press 1:EDIT if there are values in L1, then left arrow to L1 column, if necessary up arrow to L1 cell, if necessary press CLEAR press ENTER if there are values in L2, then up arrow to L1 cell right arrow to L2 cell press CLEAR press ENTER if necessary, use arrow keys to get to line 1 of L1 1 ENTER 3 ENTER right arrow 5 ENTER 15 ENTER There are now two point entered. L1 | L2 -------------- 1 | 5 3 | 15 press STAT right arrow to CALC press 4:LinReg(ax + b) home screen says: LinReg(ax + b) [cursor] press 2ND 1 (L1) press COMMA (,) press 2ND 2 press COMMA (,) press VARS right arrow to Y-VARS press 1:Function press 1:Y1 home screen says: LinReg(ax + b) L1,L2,Y1 [cursor] press ENTER home screen says: LinReg y = ax+b a = 5 b = 0 r^2 = 1 r = 1 press 2ND GRAPH to see a table press GRAPH to see a graph of the line press Y= to see the linear equation

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