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Do Now. 1. If f(x) = │(x² – 50)│, what is the value of f(-5) ? A. 75 B. 25 C. 0 D. -25 E. -75 2. ( √2 - √3 )² = A. 5 - 2√6 B. 5 - √6 C. 1 - 2√6 D. 1 - √2 E. 1 3. 2 30 + 2 30 + 2 30 + 2 30 = A. 8 120 B. 8 30 C. 2 32 D. 2 30 E. 2 26. P.3 Lines in the Plane.
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Do Now 1.If f(x) = │(x² – 50)│, what is the value of f(-5) ? A. 75 B. 25 C. 0 D. -25 E. -75 2.( √2 - √3 )² = A. 5 - 2√6 B. 5 - √6 C. 1 - 2√6 D. 1 - √2 E. 1 3.230 + 230 + 230 + 230 = A. 8120B. 830C. 232D. 230E. 226
P.3 Lines in the Plane SWBAT 1. Find the slope of a line. 2. Write linear equations given points on lines and their slopes. 3. Use slope-intercept forms of linear equations to graph lines. 4. Use slope to identify parallel and perpendicular lines.
The slope of a line • Slope: rise over run, the change in y over the change in x
The slope of a line • Finding slope given 2 points: • 1. (-2, 0) and (3, 1) • 2. (-1, 2) and (2, 2) • 3. (3, 4) and (3, 1)
The point-slope form of a line • Point-slope form: y – y1 = m(x – x1) • Find an equation of the line that passes through the point (1, -2) and has a slope of 3. • Find an equation of the line that passes through (2, -3) and (4, 5).
Using point-slope to create linear models • During 1997, Barnes and Noble’s net sales were $2.8 billion, and in 1998 net sales were $3.0 billion. • 1. Write a linear equation giving the net sales y in terms of the year x. • 2. Use the equation to estimate the net sales during 2000.
Linear extrapolation: an approximation in which the estimated point lies outside of the given points • Linear interpolation: the estimated point lies between the two given points
Sketching graphs of lines • Slope-intercept form: y = mx + b • Better suited to graph linear equations
Sketching graphs of lines • Determine the slope and y-intercept of each linear equation and describe its graph. • 1. x + y = 2 • 2. y = 2
Parallel Lines • Parallel lines: have slopes that are equal • Find the slope-intercept form of the equation of the line that passes through the point (2, -1) and is parallel to the line 2x – 3y = 5.
Perpendicular Lines • Perpendicular lines: have slopes that are negative reciprocals • Find an equation of the line that passes through the point (2, -1) and is perpendicular to the line 2x – 3y = 5.