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Day 5 β Forms of Equation
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Use graphics calculator to graph the equation 4π₯+2π¦=588.
Example 1 Use graphics calculator to graph the equation 4π₯+2π¦=588.
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Answer To enter an equation in a graphics calculator, first rewrite the equation in slope-intercept form, π¦=ππ₯+π.
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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, π΄π₯+π΅π¦=πΆ, the slope is β π΄ π΅ and the y-intercept is πΆ π΅ . Explain why.
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Solving π΄π₯+π΅π¦βπΆ for y gives the slope-intercept form π¦=β π΄ π΅ π₯+ πΆ π΅ .
Answer Solving π΄π₯+π΅π¦βπΆ for y gives the slope-intercept form π¦=β π΄ π΅ π₯+ πΆ π΅ .
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Point-Slope Form The form π¦β π¦ 1 =π(π₯β π₯ 1 ) is the point-slope form for the equation of a line. The coordinates (π₯ 1 , π¦ 1 ), and the slope is m.
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Example 2 A line with slope 3 passes through point (2, 7). Write the equation of the line in point-slope form.
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Answer Let π=3, π₯ 1 =2, πππ π¦ 1 =7. Substitute the given values into the point-slope equation. π¦β π¦ 1 =π(π₯β π₯ 1 ) π¦β7=3(π₯β2)
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Example 3 Janetβs class is ordering T-shirts. Write the point-slope equation that models the information on the order form.
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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 π₯ 1 , and 149 for π¦ 1 into π¦β π¦ 1 = π(π₯β π₯ 2 ). The result is π¦β149=9 π₯β16 .
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Example 4 Change π¦β149=9(π₯β16) to and equation a) In slope-intercept form. b) In standard form.
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Answer a) The equation is now in slope-intercept form. Note that the shipping and handling charge is $5.
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b) Next, change the slope-intercept form to a standard form, π΄π₯+π΅π¦=πΆ.
Answer b) Next, change the slope-intercept form to a standard form, π΄π₯+π΅π¦=πΆ.
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Example 5 Find an equation in point-slope form for the graph of a line that passes thought the points (β1, 10) and (5, 8).
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Answer Let π₯ 1 , π¦ 1 =(β1, 10) and π₯ 2 , π¦ 2 =(5, 8). Find the slope. πππππ= π¦ 2 β π¦ 1 π₯ 2 β π₯ 1 = 8β10 5β(β1) = β2 6 =β 1 3 Use the point (β1, 10)for( π₯ 1 β π¦ 1 ). Substitute the slope and the coordinates of the point into the equation π¦β π¦ 1 =π(π₯β π₯ 1 ). Then π¦β10=β 1 3 π₯β β1 , or π¦β10=β 1 3 (π₯+1).
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summary of the forms for linear equations
Name Form Example Slope-intercept π¦=ππ₯+π π¦=3π₯+5 Standard π΄π₯+π΅π¦=πΆ 3π₯βπ¦=β5 Point-Slope π¦β π¦ 1 =π(π₯β π₯ 1 ) π¦β11=3(π₯β2)
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