FINDING THE EQUATION OF A TANGENT LINE

Let f be a function defined on $\mathbb{R}$ to $\mathbb{R}$ and C be the curve whose equation is y = f (x). This article shows us how to determine the equation of the tangent line to C at the point K(a,b) where b = f (a). Look at the figure below.

Figure 1

The equation of the tangent line is determined by the following formula:

$y - b = f^{\: \prime} (a) (x - a)$ …………………………………………………………………………………………………………………………………………………………………… (*)

where f ‘(a) is the derivative of f at x = a.

Example 1

Find the equation of the tangent line to the curve given by y = x³ – 2x² at the point K(1,-1).

Answer

In this example, $f(x) = x^3 - 2x^2$ so $f^{\: \prime} (x) = 3x^2 - 4x$ . Substituting x = 1 (i.e. the abscissa of K) into $f^{\: \prime} (x)$ , we get $f^{\: \prime} (1) = 3 \cdot 1^2 - 4 \cdot 1 = -1$ . By substituting a = 1, b = -1, and $f^{\: \prime} (1) = -1$ into (*), we have:

y – (-1) = (-1)⋅(x – 1)

y + 1 = -x + 1

y = -x

So, the equation of the tangent line to the curve at K is y = -x. (See Figure 2.)

Figure 2

Example 2

Find the equation of the tangent line to the curve with equation y = 3x⁵ – 5x³ at point K(-1,2).

Answer

In this example, $f(x) = 3x^5 - 5x^3$ so $f^{\: \prime} (x) = 15x^4 - 15x^2$ . Substituting x = -1 (i.e. the abscissa of K) into $f^{\: \prime} (x)$ , we get $f^{\: \prime} (-1) = 15 \cdot (-1)^4 - 15(-1)^2 = 0$ . By substituting a = -1 , b = 2, and $f^{\: \prime} (-1) = 0$ into (*), we get:

y – 2 = 0(x – (-1))

y – 2 = 0

y = 2

So, the equation of the tangent line to the curve at K is y = 2. (See Figure 3.)

Figure 3

Example 3

Find the equation of the tangent to the curve with equation $y = 2 \sqrt{x}$ at x = 4.

Answer

In this example, $f(x) = 2 \sqrt{x} = 2x^{1/2}$ . Then $f^{\: \prime} (x) = x^{-1/2}$ . Substituting x = 4 into the $f^{\: \prime} (x)$ , we get $f^{\: \prime} (4) = 4^{-1/2} = \frac{1}{2}$ . To determine the ordinate of the point whose abscissa is 4, substitute x = 4 into $y = 2 \sqrt{x}$ . This gives y = 4. The equation of the tangent line to the curve at the point (4,4) will be obtained by substituting a = 4, b = 4, and $f^{\: \prime} (4) = \frac{1}{2}$ into (*). This yields:

$y - 4 = \frac{1}{2}(x - 4)$

$y = \frac{1}{2}x + 2$

Thus, the equation of the tangent line to the curve at x = 4 is $y = \frac{1}{2}x + 2$ . (See Figure 4.)

Figure 4

Example 4

Find the equation of the tangent line to the curve with the equation y = 4 sin x at the point whose abscissa is π/3.

Answer

In this example, f(x) = 4 sin x so f ‘(x) = 4 cos x. Substituting x = π/3 into the f ‘(x), we have f ‘(π/3) = 4 cos π/3 = 4⋅½ = 2. To determine the ordinate of the point whose abscissa is π/3, substitute x = π/3 into y = 4 sin x. This gives $y = 4 \cdot \frac{1}{2} \sqrt{3} = 2 \sqrt{3}$ . The equation of the tangent line to the curve at the point $(\frac{\pi}{3},2 \sqrt{3})$ will be obtained by substituting a = π/3, b = 2√3, and f ‘(π/3) = 2 into (*). This results in: