1. 1 Example 6 Sketch the graph of the function Solution I. Intercepts The x-intercepts occur when the numerator of q(x) is zero i.e. when x=1. The y-intercept occurs at II. Asymptotes Vertical asymptotes occur where the denominator of q(x) is zero, i.e at x=8. The horizontal asymptote on the right is given by: The horizontal asymptote on the left is given by: Thus q has the horizontal asymptote y=1 on both the left and the right.

2. 2 III. First Derivative By the chain and quotient rules, the derivative of is: The sign of q / (x) is determined by the sign of its denominator. Hence q / (x) is positive for 1 8 where q is decreasing. We depict this information on a number line. q has only one critical point: x=1 where q / (x) does not exist. By the First Derivative Test, x=1 is a local minimum Note that x=8 is not a critical point of q because q has a vertical asymptote there and x=8 not in the domain of q.

3. 3 IV. Vertical Tangents and Cusps q has a vertical cusp at x=1 because the left derivative is -  while the right derivative is + . V. Concavity and Inflection Points By the product and chain rules, the derivative of is: Observe that the denominator of q // (x) is always positive. Hence q // (x) has the same sign as its numerator 84x-182. Thus q // (x)>0 for x>91/42 and q is concave up there while q // (x)<0 for x<91/42 and q is concave down there. We depict this information on a number line.

4. 4 Since the concavity of q changes from down to up at x=91/42, the function q has an inflection point there. VI. Sketch of the graph We summarize our conclusions and sketch the graph of q.

5. 5 x-intercepts: x=1 y-intercept: y=1/4 vertical asymptote: x=8 horizontal asymptote: y=1 on the left and right vertical cusp: (1,0) decreasing: x<1 or 8<x increasing: 1<x<8 local min: x=1 concave down: x 91/42 inflection point: x=91/42