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# Tangent Lines for AP Calculus

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By McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: More Applications of Derivatives Practice Problems for AP Calculus

If the function y is differentiable at x = a, then the slope of the tangent line to the graph of y at x = a is given as m(tangent at x = a) =

### Types of Tangent Lines

Horizontal Tangents (See Figure 9.1-1.)

Vertical Tangents . (See Figure 9.1-2.)

Parallel Tangents (See Figure 9.1-3.)

### Example 1

Write an equation of the line tangent to the graph of y = – 3 sin 2x at x = . (See Figure 9.1-4 on page 178.)

y = – 3 sin 2x; = – 3[cos(2x )]2 = – 6 cos(2x )

Slope of tangent = – 6 cos [2(π/2)]= – 6 cosπ = 6.

Point of tangency: At x = , y = – 3 sin(2x )

= – 3 sin[2(π/2)]= – 3 sin(π)=0.

Therefore, is the point of tangency.

Equation of Tangent: y – 0 = 6(x – π/2) or y = 6x – 3π.

### Example 2

If the line y =6x + a is tangent to the graph of y =2x3, find the value(s) of a.

Solution:

y = 2x3; = 6x2. (See Figure 9.1-5.)

The slope of the line y = 6x + a is 6.

Since y = 6x + a is tangent to the graph of x = 2x3, thus = 6 for some values of x.

Set 6x2 = 6 x2 =1 or x = ± 1.

At x = –1, y = 2x3 = 2(–1)3 = –2; (–1, –2) is a tangent point. Thus, y = 6x + a –2= 6(–1)+ a or a = 4.

Therefore, a = ± 4.

### Example 3

Find the coordinates of each point on the graph of y2x2 – 6x + 7 = 0 at which the tangent line is vertical. Write an equation of each vertical tangent. (See Figure 9.1-6.)

Step 1:  Find .

Step 2:  Find

Step 3:  Find points of tangency.

At y = 0, y2x2 – 6x +7 = 0 becomes –x2 – 6x +7 = 0 x2 +6x – 7 = 0 (x +7)(x – 1) = 0 x = – 7 or x =1.

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

Step 4:  Write equation for vertical tangents:

x = – 7 and x = 1.

### Example 4

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

Step 1:  Find

Step 2:  Find the x -coordinate of points of tangency.

Horizontal Tangent = 0.

If x ≥ 0, set e x + xex = 0 ex (1+ x ) = 0 x = – 1 but x ≥ 0, therefore, no solution.

If x < 0, set –exxex = 0 ex(1+ x ) = 0 x = – 1.

Step 3:– Find points of tangency.

At x = – 1, y = – xex = – (–1)e–1 = .

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

### Example 5

Using your calculator, find the value(s) of x to the nearest hundredth at which the slope of the line tangent to the graph of y = 2 ln(x2 + 3) is equal to – . (See Figures 9.1-8 and 9.1-9.)

Step 1:– Enter y 1=2 * ln (x^2+3)

Step 2:– Enter y2 = d(y1(x ), x ) and enter y3 = –

Step 3:– Using the [Intersection] function of the calculator for y2 and y3, you obtain x = – 7.61 or x = – 0.39.

### Example 6

Using your calculator, find the value(s) of x at which the graphs of y = 2x2 and y = ex have parallel tangents.

Step 1 : Find for both y = 2x2 and y = ex.

Step 2:  Find the x -coordinate of the points of tangency. Parallel tangents slopes are equal.

Set 4x = ex 4xex = 0

Using the [Solve] function of the calculator, enter [Solve] (4xe^(x )= 0, x ) and obtain x = 2.15 and x =0.36.

Practice problems for these concepts can be found at:

More Applications of Derivatives Practice Problems for AP Calculus

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