5 Steps to a 5 AP Calculus BC 2019

178 STEP 4. Review the Knowledge You Need to Score High

Step 3: Find points of tangency.

Aty=0,y^2 −x^2 − 6 x+ 7 =0 becomes−x^2 − 6 x+ 7 = 0 ⇒ x^2 + 6 x− 7 = 0

⇒(x+7)(x−1)= 0 ⇒x=−7orx=1.

Thus, the points of tangency are (−7, 0) and (1, 0).

Step 4: Write equation for vertical tangents.

x=−7 andx=1.

Example 4

Find all points on the graph ofy=|xex|at which the graph has a horizontal tangent.

Step 1: Find

dy

dx

.

y=|xex|=

{

xex ifx≥ 0

−xex ifx< 0

dy

dx

=

{

ex+xex ifx≥ 0

−ex−xex ifx< 0

Step 2: Find thex-coordinate of points of tangency.

Horizontal tangent⇒

dy

dx

= 0.

Ifx ≥0, setex+xex= 0 ⇒ex(1+x)= 0 ⇒x=−1 butx≥0, therefore, no

solution.

Ifx<0, set−ex−xex= 0 ⇒−ex(1+x)= 0 ⇒x=−1.

Step 3: Find points of tangency.

Atx=−1,y=−xex=−(−1)e−^1 =

1

e

.

Thus at the point (−1, 1/e), the graph has a horizontal tangent. (See Figure 9.1-7.)

[−3, 1] by [−0.5, 1.25]

Figure 9.1-7