2.2 graphs of first degree functions t
- 1. a.2x – 3y = 12 b. –3y = 12 c. 2x = 12 If both x and y are present, we get a tilted line. If the equation is y = c, we get a horizontal line. The graphs of the equations Ax + By = C are straight lines. It's easy to graph lines by graphing the x and y intercepts, i.e. set x = 0 to get the y–intercept, and set y = 0 for the x–intercept. If there is only one variable in the equation, we get a vertical or a horizontal line. (6,0) (0,–4) y = –4 x = 6 If the equation is x = c, we get a vertical line. First Degree Functions
- 2. Linear Equations and Lines Example A. A river floods regularly, and on a rock by the river there is a mark indicating the highest point the water level ever recorded. At 12 pm July 11, the water level is 28 inches below this mark. At 8 am July 12 the water is 18 inches below this mark. Let x = time, y = distance between the water level and the mark. Find the linear function y = f(x) = mx + b of the distance y in terms of time x. Since how the time was measured is not specified, we may select the stating time 0 to be time of the first observation. By setting x = 0 (hr) at 12 pm July 11, then x = 20 at 8 am of July 12.
- 3. Using the point (0, 28) and the point–slope formula, y = – ½ (x – 0) + 28 or Δy Δx 28 – 18 0 – 20 Equations of Lines In particular, we are given that at x = 0 →y = 28, and at x = 20 → y = 18 and that we want the equation y = m(x – x1) + y1 of the line that contains the points (0, 28) and (20, 18). =The slope m = = –1/2 – xy = + 28 2 The linear equation that we found is also called the trend line. So if at 4 pm July 12, i.e. when x = 28, we measured that y = 12” but based on the formula prediction that y should be – 28/2 + 28 = 14”, we may conclude that the flood is intensifying.
- 4. More Facts on Slopes: • Parallel lines have the same slope. • Slopes of perpendicular lines are the negative reciprocal of each other. Example B. Find the equation of the line L that passes through (2, –4) and is perpendicular to 4x – 3y = 5. Solve for y to find the slope of 4x – 3y = 5 4x – 5 = 3y 4x/3 – 5/3 = y Hence the slope of 4x – 2y = 5 is 4/3. Therefore L has slope –3/4. So the equation of L is First Degree Functions y = (–3/4)(x – 2) + (–4) or y = –3x/4 – 5/2.