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# Graphing and Writing Linear Functions

## Linear Functions

Definition: A l i n e a r equation can be written in the form Ax + By = C.

Note: These equations are called linear because their graphs are always lines.

Question: Which linear equations are functions?

Answer: All lines except vertical lines are functions.

## Slope-Intercept Form

Definition: A l i n e a r function can be written in the form f(x) = mx + b. This is also referred to as the slope-intercept form of a line, where m is the slope and b is the y-intercept.

Note: The slope is the ratio (or comparison) of the change in y divided by the change in x. A good way to remember slope is to
use .

## Graphing Linear Functions

To graph a linear function we need to recall a fact from geometry. We will use the fact that any two points define a line. Thus to graph a linear function we need only find two solution points to the linear equation. There are many ways to accomplish this.

Method 1: Use any Two Points

Example1: f(x) = 2 x - 3

Solution: We need to find two solution points.  Example 2: Solution:  Method 2: Use the Intercepts

The intercepts are the points where the graph crosses each axis. There are x-intercepts and y-intercepts. We find the x-intercepts by letting y = 0 and the y-intercepts by letting x = 0.

Example 1: h(x) = -3 x - 6

Solution: The y-intercept is (0, -6).

To get the x-intercept let y = 0.
0 = -3 x - 6
6 = -3 x
x = -2

The x-intercept is(-2, 0). Example 2: Solution: The y-intercept is (0, 4).
To get the x-intercept. So the x-intercept is (-6, 0). Method 3: Use the Slope and y-intercept

We can use the slope and intercept of the line to sketch it. Start by plotting the y-intercept and then use the slope as a map to find a second point.

Example 1: s(x) =-4 x + 1

Solution: Start with the y-intercept (0, 1), them from move down 4 and right 1. This gives the point (1,-3). Example 2: Solution: ## Writing Linear Functions

To write the linear function that passes through two points, we'll use the slope-intercept form f(x) = m x + b. This means we must first find the slope of the line and then find the y-intercept.

Example 1: Find the equation of the linear function passing through the two points (2,1) and (0,3). Then write the equation
using function notation.

Solution: First we need to find the slope. So this tells us that the function is of the form y = -x + b.
To get b, we'll note that the point (0, 3) is the y-intercept and so b = 3.
So our linear function is f(x) = -x + 3

Example 2: Find the equation of the linear function passing through the two points (3,6) and (-3,2). Then write the equation
using function notation.

Solution: So this tells us that the function is of the form .
To find b, take one of the points and substitute and then solve. So the linear function is Note: There is an alternate way to find the equation of a line. We can use point-slope form of a line. y - y1 = m(x - x1)

Example 3: Find the equation of the linear function passing through the two points (5, -3) and (-10, 3). Then write the
equation using function notation.

Solution: .
So the equation is given by .

Now to get this into function form we need to solve for y and simplify. ## Parallel and Perpendicular Lines

Theorem:
Two lines are parallel if they have equal slopes.
Two lines are perpendicular if they have opposite reciprocal slopes.
i.e. If the first line has slope , then the second has slope - .

Example 1: Find the equation of the line passing through the point (1,5) and parallel to the line 4 x + 2 y = 1.

Solution: First we need to find the slope. Since the new is parallel to 4 x + 2 y = 1, we need to find the slope of this line. We can do this by solving for y and getting slope-intercept form. So m = -2.
So our new line is of the form y = -2 x + b.

5 = -2(1) + b
b = 7

So the new linear function is f(x) = -2 x + 7.

Example 2: Find the equation of the line passing through the point (2,1) and perpendicular to the line 2 x + 3 y = 15.

Solution: We need to find the slope of 2 x + 3 y = 15. So m = , since the new line is perpendicular. So it is of the form  So the new function is .