Day 5 – Forms of Equation

Use graphics calculator to graph the equation 4𝑥+2𝑦=588. Example 1 Use graphics calculator to graph the equation 4𝑥+2𝑦=588.

Answer To enter an equation in a graphics calculator, first rewrite the equation in slope-intercept form, 𝑦=𝑚𝑥+𝑏.

Explain why. The equation can now be entered into the calculator. Answer The equation can now be entered into the calculator. If the equation of a line is written in standard form, 𝐴𝑥+𝐵𝑦=𝐶, the slope is − 𝐴 𝐵 and the y-intercept is 𝐶 𝐵 . Explain why.

Solving 𝐴𝑥+𝐵𝑦−𝐶 for y gives the slope-intercept form 𝑦=− 𝐴 𝐵 𝑥+ 𝐶 𝐵 . Answer Solving 𝐴𝑥+𝐵𝑦−𝐶 for y gives the slope-intercept form 𝑦=− 𝐴 𝐵 𝑥+ 𝐶 𝐵 .

Point-Slope Form The form 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) is the point-slope form for the equation of a line. The coordinates (𝑥 1 , 𝑦 1 ), and the slope is m.

Example 2 A line with slope 3 passes through point (2, 7). Write the equation of the line in point-slope form.

Answer Let 𝑚=3, 𝑥 1 =2, 𝑎𝑛𝑑 𝑦 1 =7. Substitute the given values into the point-slope equation. 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑦−7=3(𝑥−2)

Example 3 Janet’s class is ordering T-shirts. Write the point-slope equation that models the information on the order form.

Answer Each time a shirt is ordered, the cost increases by $9, so 9 is the rate of change, or slope. Since 16 shirts cost $149, the point (16, 149) represents a point on the graph. Substitute 9 for m, 16 for 𝑥 1 , and 149 for 𝑦 1 into 𝑦− 𝑦 1 = 𝑚(𝑥− 𝑥 2 ). The result is 𝑦−149=9 𝑥−16 .

Example 4 Change 𝑦−149=9(𝑥−16) to and equation a) In slope-intercept form. b) In standard form.

Answer a) The equation is now in slope-intercept form. Note that the shipping and handling charge is $5.

b) Next, change the slope-intercept form to a standard form, 𝐴𝑥+𝐵𝑦=𝐶. Answer b) Next, change the slope-intercept form to a standard form, 𝐴𝑥+𝐵𝑦=𝐶.

Example 5 Find an equation in point-slope form for the graph of a line that passes thought the points (−1, 10) and (5, 8).

Answer Let 𝑥 1 , 𝑦 1 =(−1, 10) and 𝑥 2 , 𝑦 2 =(5, 8). Find the slope. 𝑆𝑙𝑜𝑝𝑒= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 8−10 5−(−1) = −2 6 =− 1 3 Use the point (−1, 10)for( 𝑥 1 − 𝑦 1 ). Substitute the slope and the coordinates of the point into the equation 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ). Then 𝑦−10=− 1 3 𝑥− −1 , or 𝑦−10=− 1 3 (𝑥+1).

summary of the forms for linear equations Name Form Example Slope-intercept 𝑦=𝑚𝑥+𝑏 𝑦=3𝑥+5 Standard 𝐴𝑥+𝐵𝑦=𝐶 3𝑥−𝑦=−5 Point-Slope 𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 ) 𝑦−11=3(𝑥−2)