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Translations of Linear Functions. Modified LTF activity. #1. Graph y = 2x. Plot at least 5 points. #1. Graph y = 2x. Plot at least 5 points On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right. #1. Graph y = 2x. Plot at least 5 points
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Translations of Linear Functions Modified LTF activity
#1 • Graph y = 2x. Plot at least 5 points
#1 • Graph y = 2x. Plot at least 5 points • On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right.
#1 • Graph y = 2x. Plot at least 5 points • On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right. • After the translation, what is the “new” location of the point (0, 0)?
#1 • Graph y = 2x. Plot at least 5 points • On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form.
#1 • Graph y = 2x. Plot at least 5 points • On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form. • We can factor out the common factor in the translated equation to write a new equation.
#1 • Graph y = 2x. Plot at least 5 points • On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form. • We can factor out the common factor in the translated equation to write a new equation. • New equation: y = 2(x – 4)
#1 • Graph y = 2x. Plot at least 5 points • On the same graph, translate the graph 4 units to the right by moving each point 4 units to the right. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form. • We can factor out the common factor in the translated equation to write a new equation. • New equation: y = 2(x – 4) • We moved the line 4 units to the right. Where does this important value appear in the new equation?
#2 • Graph y = 2x again. Plot at least 5 points
#2 • Graph y = 2x again. Plot at least 5 points • On the same graph, translate the graph 4 units to the left by moving each point 4 units to the left.
#2 • Graph y = 2x again. Plot at least 5 points • On the same graph, translate the graph 4 units to the left by moving each point 4 units to the left. • After the translation, what is the “new” location of the point (0, 0)?
#2 • Graph y = 2x again. Plot at least 5 points • On the same graph, translate the graph 4 units to the left by moving each point 4 units to the left. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form.
#2 • Graph y = 2x again. Plot at least 5 points • On the same graph, translate the graph 4 units to the left by moving each point 4 units to the left. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form. • We can factor out the common factor in the translated equation to write a new equation.
#2 • Graph y = 2x again. Plot at least 5 points • On the same graph, translate the graph 4 units to the left by moving each point 4 units to the left. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form. • We can factor out the common factor in the translated equation to write a new equation. • New equation: y = 2(x + 4)
#2 • Graph y = 2x again. Plot at least 5 points • On the same graph, translate the graph 4 units to the left by moving each point 4 units to the left. • After the translation, what is the “new” location of the point (0, 0)? • Write the equation of the translated line in slope-intercept form. • We can factor out the common factor in the translated equation to write a new equation. • New equation: y = 2(x + 4) • We moved the line 4 units to the left. Where does this important value appear in the new equation?
#3 Answer these reflection questions • For #1 and #2, the only difference was the direction the graphs moved. What is the only difference in the two new equations? • Did addition correspond to moving left or right? What about subtraction? Is that what you would have expected? • What is the significance of the common factor (the number outside the parenthesis)?
#5 • Graph y = ¼ x. Plot at least 4 points.
#5 • Graph y = ¼ x. Plot at least 4 points. • On the same graph, translate the line to the right 2 units and up 3 units.
#5 • Graph y = ¼ x. Plot at least 4 points. • On the same graph, translate the line to the right 2 units and up 3 units. • We know that the horizontal translation is an x movement and shows up as y = ¼ (x – 2). The vertical translation will appear next to y. Do you think it will be y + 3 or y – 3?
#5 • Graph y = ¼ x. Plot at least 4 points. • On the same graph, translate the line to the right 2 units and up 3 units. • We know that the horizontal translation is an x movement and shows up as y = ¼ (x – 2). The vertical translation will appear next to y. Do you think it will be y + 3 or y – 3? • Point-Slope Equation: y – 3 = ¼ (x – 2)
#5 • Graph y = ¼ x. Plot at least 4 points. • On the same graph, translate the line to the right 2 units and up 3 units. • We know that the horizontal translation is an x movement and shows up as y = ¼ (x – 2). The vertical translation will appear next to y. Do you think it will be y + 3 or y – 3? • Point-Slope Equation: y – 3 = ¼ (x – 2) • The “new” (0, 0) is now at (2, 3). How does this point show up in the equation?
#6 – Try this one on your own! • Graph y = 3x. Plot at least 4 points. • Translate the line 1 unit left and 3 units down. • Write the point-slope equation of the translated line. • What is the slope of this line? What point do we know for sure is on the line based on the equation?
Notes • This equation that shows the translation is written as y – y1 = m(x – x1) and is called POINT-SLOPE FORM. • The m stands for slope. The (x1 , y1 ) represents one point on the line. • When we use point-slope form in application, we no longer care about the “original” line and the translation. Instead we want to make sure we can name the slope and one point.
#9 – Can you write it from this format? GIVEN TWO POINTS: (4, 5) AND (3, -2) For each format, you need to find slope and choose one point. This is enough information to write the point-slope equation.