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Gold Day – 2/26/2015 Blue Day – 3/2/2015.  Unit 5 – Linear functions and Applications  Standard Form  Finding equations of a line using the intercepts.

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Presentation on theme: "Gold Day – 2/26/2015 Blue Day – 3/2/2015.  Unit 5 – Linear functions and Applications  Standard Form  Finding equations of a line using the intercepts."— Presentation transcript:

1 Gold Day – 2/26/2015 Blue Day – 3/2/2015

2  Unit 5 – Linear functions and Applications  Standard Form  Finding equations of a line using the intercepts  Point-Slope form

3 Find the slope. 1. (2, -5) (3, 1) 2. (-6, 4) (3, 7) 3. (9,0) (0,9) 4. (-1,4) (-2,-5) 2 1 1 2 Hint: Use either formula

4  Change Ax + By=C to y=mx + b Ex: 1. -3x + 4y = 82. 6x + 3y = 27

5  Change y=mx + b to Ax + By=C Ex: 1. y=x + 22. y=- x + 9

6  So we have talked about linear functions being represented in a couple of forms.  Slope-intercept form ▪ Provides an m(slope) & b(y-intercept) for graphing the line. ▪ We did this on calculator and by hand.  Standard form ▪ Can replace the x and y with zeros to find the x- and y- intercepts for graphing the line  Now…point-slope form ▪ Provides 1 point and m(slope) to find the graph of the line.

7 Point- Slope form is: Where m is the slope and (x 1, y 1 ) is the given point

8 y – y 1 = m(x – x 1 ) (x 1,y 1 )m = rise = slope point run

9  Given a point and a slope, write an equation in point-slope form y – y 1 = m(x – x 1 ) Example 1: (3,8) m = 2 y – 8 = 2 (x – 3) x 1, y 1

10 y – y 1 = m(x – x 1 ) (-6,1) m = -4 y – 1 = -4(x – (-6)) y – 1 = -4(x + 6) x 1, y 1

11 y – y 1 = m(x – x 1 ) (5,-2) m= x 1, y 1

12 Given 2 ordered pairs, write an equation in point-slope form: (6,3) (-8,5) Steps: 1. Find slope m = y 2 – y 1 x 2 – x 1 2. Pick an ordered pair There are 2 possible answers!

13 Given 2 ordered pairs, write an equation in point-slope form: (-1,-7) (1,3)

14 Given 2 ordered pairs, write an equation in point-slope form: (-4,6) (-2,5)

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