1.4-5 Linear Functions and Slope
Slope: the slope m of a non-vertical line is the number of units the line rises or falls for each unit of
horizontal change from left to right
𝑚=
𝑦2 − 𝑦1
𝑟𝑖𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠
=
=
𝑥2 − 𝑥1
𝑟𝑢𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠
Example: Find the slope between the pair of coordinates (-3, 2) and (1, 5)
5−2
3
𝑚 = 1−−3 = 4
Try these: (-4, 0) and (3, 3)
(-1, -2) and (2, -6)
(-3, -1) and (1, 3)
x-intercept: the x-coordinate of a point where a graph crosses the x-axis. The y-coordinate of the point
is 0
Example: Find the x-intercept of 4x – 2y = 8
Let y=0
4x – 2(0) = 8 and solve for x…..x = 2
The x-intercept is (2, 0)
Try these: x – y = 6
-2x + y = -4
3x – 2y = 6
y-intercept: the y-coordinate of a point where a graph crosses the y-axis. The x-coordinate of the point
is 0.
Example: Find the y-intercept of 4x-2y = 8
Let x=0
4(0) – 2y = 8 and solve for y………..y= -4
The y-intercept is (0, -4)
Try these: x – y = 6
-2x + y = -4
Slope-Intercept Form: y = mx + b
3x – 2y = 6
where m = slope, and b = y-intercept
Example: Find the slope and y-intercept of y = -5x -1
m = -5, y-int= (0, -1)
Try: y = -x -1
5x + 2y = 6 solve for y first to get it in y = mx + b form!!
Example: Find the equation that has a slope m=2 and y-intercept (0, -5)
y = mx + b = 2x – 5
y = 2x – 5
TRY: Find the equation that has a slope of -1 and y-intercept (0, -6)
How to graph Slope-Intercept Form
1.
2.
3.
4.
First, write the equation in y = mx + b form.
Determine your slope and your y-intercept.
Plot your y-intercept.
Draw a slope triangle to locate a second point on the line.
𝑟𝑖𝑠𝑒
a. A slope triangle can be created using your slope. Rewrite it as 𝑟𝑢𝑛 from your
point you will go over the run and up the rise.
5. Draw a line through the two points.
Example: Graph 5x – y = 3
1. y = 5x – 3
2. m = 5 y-int = (0, -3)
3. Plot y-int
5
4. m= 5=1 we are going to go up 5 and over 1
5. Connect the points.
Try this: Graph 6x – y = 0
Point-Slope Form: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) where (𝑥1 , 𝑦1 ) is a point, m is your slope, and x and y are
variables.
Example: Find the equation of the straight line that has slope m = 4 and passes through the
point (-1, -6).
Plug in everything you know
y – (-6) = 4(x – (-1))
Simplify and solve for y.
y + 6 = 4 (x + 1)
y + 6 = 4x + 4
y = 4x -2
TRY: Find the equation of the straight line that has slope m = -2 and passes through the point (2,
-3)
Example: Find the equation of the line that passes through the points (-2, 4) and (1, 2)
First- find the slope
4−2
𝑚 = −2−1 =
2
−3
2
= −3
Second- now plug in one of the given points and your slope in to the equation and solve
as we did before.
2
y – 4 = (- 3) (x – (-2))
2
y – 4 = (- 3) (x + 2)
2
4
y – 4 = (- 3)x – 3
2
8
y = (- 3)x + 3
TRY: Find the equation of the line that passes through the points (2, 3) and (0, -4)
NOTE: The generic point-slope form WILL be accepted as an answer!!
It does not need to be simplified to slope-intercept form!!!!
Parallel Lines: lines are parallel to each other if they have an identical slope!
Example: Find a line that is parallel to 2x – 3y = 9 and passes through the point (4, -1)
Find the slope of the line—simplify into slope-intercept form
2x – 3y = 9
-3y = -2x + 9
2
3
y= x–3
2
slope= m = 3
Now we can use the point-slope form to find the equation of the parallel line
2
y – (-1) = 3(x - 4)
2
8
y + 1 = 3x - 3
2
3
y= x-
11
3
Try: Find a line that is parallel to 5x – y = 4 and passes through the point (-4, -1)
Perpendicular Lines: lines are perpendicular if their slopes are negative reciprocals. What this means is
that you flip their slope and negate it. A slope of m= 4 would have a negative reciprocal of -1/4
Example: Find a line that is perpendicular to 2x – 3y = 9 and passes through the point (4, -1)
2
We found earlier that the slope of this is m= 3 now we need to find the negative reciprocal of
3
this so we flip it and negate it so our perpendicular slope is m= − 2
Now we plug as we did before to our point-slope form
3
y – (-1) = − 2(x – 4)
3
2
y=− x+5
TRY: Find a line that is perpendicular to 5x – y = 4 and passes through the point (-4, -1)
Average Rate of Change of a Function
Δ𝑦
Δ𝑥
=
𝑓(𝑥2 )−𝑓(𝑥1 )
𝑥2 −𝑥1
Example: f(x)= x²+2 from x = 1 to x = 3
Δ𝑦
Δ𝑥
=
𝑓(3)−𝑓(1)
3−1
=
11−3
2
8
=2=4
Average Velocity of a Function
∆𝑠
∆𝑡
=
𝑠(𝑡2 )−𝑠(𝑡1 )
𝑡2 −𝑡1
Example: the distance of a ball is modeled by s(t)=4t². What is the average velocity from t = 1 to
t=2?
∆𝑠
∆𝑡
=
𝑠(2)−𝑠(1)
2−1
=
16−4
1
= 12