The equations are written out carefully in 12.3.3.Topic A builds on work from Unit 3 to expand the idea of a solution to a coordinate point and to review identifying features of linear functions as well as graphing and writing equations in different forms to reveal properties.Question 16: Write the equation of a line in slope-intercept form with a slope of 4/3 that goes through the point (-9, 3). Numerical techniques used for evaluating these integrals are outlined. These expressions involve infinite integrals over wave number, sometimes known as Sommerfield integrals. Equations are given for the field of a point source in a homogeneous or stratified earth.This shows that the line x = -2 is a vertical asymptote for the graph of f. (The numerator approaches 36 and the denominator is a positive number approaching 0. (The numerator approaches 36 and the denominator is a negative number approaching 0. Remember, if EITHER of these one-sided limits isĪ vertical asymptote exists. This shows that the line x = 2 is a vertical asymptote for the graph of f. (The numerator approaches 4 and the denominator is a negative number approaching 0. (The numerator approaches 4 and the denominator is a positive number approaching 0. Now check for vertical asymptotes by computing one-sided limits at the zeroes of the denominator, i.e., at x=2 and at x=-2. Thus, the line y=1 is a horizontal asymptote for the graph of f. If x=0, then y=-4 so that y=-4 is the y-intercept. See the adjoining sign chart for the second derivative, f''. However, note that f'' is NOT DEFINED at x=2. Now determine a sign chart for the second derivative, f''. See the adjoining sign chart for the first derivative, f'. In addition, note that f' is NOT DEFINED at x=2. Using the quotient rule, we getįor x= 1, and x=3. Now determine a sign chart for the first derivative, f'. SOLUTION 5 : The domain of f is all x-values EXCEPT x=2, because of division by zero. Ĭlick HERE to return to the list of problems. Thus, the line y = 0 is a a horizontal asymptote for the graph of f. So that 4 x=0 and x=0 is the x-intercept. If x=0, then y=0 so that y=0 is the y-intercept. Now summarize the information from each sign chart.į has an absolute minimum at x=-1, y=-2.

Now determine a sign chart for the second derivative, f'' :įor x=1. Now determine a sign chart for the first derivative, f' :įor x=0 and x=2. SOLUTION 1 : The domain of f is all x-values. SOLUTIONS TO GRAPHINGOF FUNCTIONS USING THE FIRST AND SECOND DERIVATIVES Solutions to Graphing Using the First and Second Derivatives

X 3x 1 3 x 1 2