Equation of a Tangent Line to a Curve

Relationship Between Derivatives and Tangent Lines

The derivative describes how a function changes at a particular input. Geometrically, that change appears as the slope of the tangent line to the graph.

If we magnify a smooth curve near one point, a sufficiently small section looks almost straight. The line that follows this local direction is the tangent line. At $x=a$ on the curve $y=f(x)$, its slope is the derivative evaluated at that input: $m=f'(a)$.

Determining the Equation of a Tangent Line

To determine a straight-line equation, we need one point on the line and its slope. For a tangent line, these are:

1. Point of Tangency: A point $P(a, b)$ where the line touches the curve.
2. Gradient (m): The slope of the line at that point, which we get from the derivative, $m = f'(a)$.

Substitute both values into the point-slope form of a line:

Differentiate the function, evaluate the derivative at the point of tangency, and use that slope with the point-slope equation.

Finding the Tangent Line to a Parabola

Determine the equation of the tangent line to the parabola $y = 2 - 4x^2$ at the point $(1, -2)$.

Solution:

Step 1: Find the gradient of the tangent line

Differentiate $f(x) = 2 - 4x^2$ to obtain the slope at any input.

The point of tangency has $x=1$, so evaluate the derivative there.

The tangent line therefore has slope $-8$.

Step 2: Construct the equation

The required point and slope are now known:

• Point of tangency $(a, b) = (1, -2)$
• Gradient $m = -8$

Substitute them into the point-slope equation:

The tangent line is therefore $y = -8x + 6$.

Exercises

Each problem asks for a tangent line that satisfies an extra condition. Use the derivative to match the required gradient before writing the equation.

1. Find the equation of the tangent line to the curve $y = x^2 - 6\frac{1}{5}x + 14\frac{1}{2}$ which is parallel to the line $x + 2y + 3 = 0$.

2. Determine the tangent line equations to the parabola $y = 2x^2 - 3x + 5$ at the points whose $y$-coordinate is $4$.

3. The curve $y = 3x - \frac{3}{x^2}$ intersects the $x$-axis at $P$. Find the tangent line equation at that point.

4. The curve $y = (x^2 + 2)^2$ intersects the $y$-axis at $A$. Show that the tangent line at $A$ is parallel to the $x$-axis and is $4 \text{ units}$ from the origin.

5. Determine the point on the curve $y = 2x^2 - 7x + 1$ where the tangent line forms an angle of $45^\circ$ with the positive $x$-axis. Then determine that tangent line equation.

Answer Key

1. Solution:

   Step 1: Determine the gradient.

   The tangent line must be parallel to $x + 2y + 3 = 0$. Convert this equation into slope-intercept form $y = mx + c$ to read its slope.

   $$2y = -x - 3$$

   $$y = -\frac{1}{2}x - \frac{3}{2}$$

   From this, we know the gradient of the line is $m = -\frac{1}{2}$. Since the tangent line is parallel, its gradient is the same.

   Step 2: Find the point of tangency.

   The curve's slope at a point equals the derivative there. Differentiate $f(x) = x^2 - \frac{31}{5}x + \frac{29}{2}$ first.

   $$f'(x) = 2x - \frac{31}{5}$$

   Next, we set this derivative equal to the known slope ($m = -\frac{1}{2}$) to find the $x$-coordinate of the point of tangency.

   $$2x - \frac{31}{5} = -\frac{1}{2}$$

   $$2x = \frac{31}{5} - \frac{1}{2} = \frac{62 - 5}{10} = \frac{57}{10}$$

   $$x = \frac{57}{20}$$

   After getting the $x$-coordinate $x=\frac{57}{20}$, we substitute this value back into the original curve equation to find its $y$-coordinate.

   $$\begin{aligned}y &= (\frac{57}{20})^2 - \frac{31}{5}(\frac{57}{20}) + \frac{29}{2} \\ &= \frac{3249}{400} - \frac{1767}{100} + \frac{29}{2} \\ &= \frac{3249 - 7068 + 5800}{400} \\ &= \frac{1981}{400}\end{aligned}$$

   The point of tangency is $\left(\frac{57}{20}, \frac{1981}{400}\right)$.

   Step 3: Construct the line equation.

   With the point $\left(\frac{57}{20}, \frac{1981}{400}\right)$ and gradient $m = -\frac{1}{2}$, the equation is:

   $$y - \frac{1981}{400} = -\frac{1}{2}\left(x - \frac{57}{20}\right)$$

   $$y = -\frac{1}{2}x + \frac{57}{40} + \frac{1981}{400}$$

   $$y = -\frac{1}{2}x + \frac{570 + 1981}{400}$$

   $$y = -\frac{1}{2}x + \frac{2551}{400}$$

2. Solution:

   Step 1: Find the point(s) of tangency.

   Since the ordinate is $4$, we set $y=4$ in the parabola's equation.

   $$4 = 2x^2 - 3x + 5$$

   $$2x^2 - 3x + 1 = 0$$

   $$(2x-1)(x-1) = 0$$

   From factorization, we get two x-values: $x=1$ and $x=\frac{1}{2}$. This means there are two points of tangency: $(1, 4)$ and $\left(\frac{1}{2}, 4\right)$. Therefore, there will be two tangent line equations.

   Step 2: Calculate the gradient and create the equation for each point.

   We will process each point of tangency separately. The derivative of the function is $f'(x) = 4x - 3$.

   Case One: Point $(1, 4)$

   The gradient at this point is $m = f'(1) = 4(1) - 3 = 1$.

   Thus, the equation is:

   $$y - 4 = 1(x - 1) \implies y = x + 3$$

   Case Two: Point $\left(\frac{1}{2}, 4\right)$

   The gradient at this point is $m = f'(\frac{1}{2}) = 4(\frac{1}{2}) - 3 = -1$.

   Thus, the equation is:

   $$\begin{aligned} y - 4 &= -1(x - \frac{1}{2}) \\ &\implies y \\ &= -x + \frac{9}{2} \end{aligned}$$

3. Solution:

   Step 1: Find point $P$.

   The curve intersects the $x$-axis when $y=0$.

   $$0 = 3x - \frac{3}{x^2}$$

   $$\begin{aligned} 3x &= \frac{3}{x^2} \\ &\implies 3x^3 \\ &= 3 \\ &\implies x^3 \\ &= 1 \\ &\implies x \\ &=1 \end{aligned}$$

   The intersection point $P$ is $(1, 0)$.

   Step 2: Find the slope at $P$.

   Rewrite the function as $f(x) = 3x - 3x^{-2}$ so the power rule can be applied directly.

   $$f'(x) = 3 - (-2)(3)x^{-3} = 3 + \frac{6}{x^3}$$

   The gradient at $x=1$ is $m = f'(1) = 3 + \frac{6}{1^3} = 9$.

   Step 3: Construct the equation.

   With point $(1, 0)$ and gradient $9$, the equation is:

   $$y - 0 = 9(x - 1) \implies y = 9x - 9$$

4. Solution:

   Step 1: Find point $A$.

   The curve intersects the $y$-axis when $x=0$.

   $$y = (0^2 + 2)^2 = 4$$

   The intersection point $A$ is $(0, 4)$.

   Step 2: Prove the tangent line is parallel to the $x$-axis.

   A line parallel to the $x$-axis has slope $0$. Evaluate the derivative at $A$, where $x=0$.

   The derivative of $f(x)=(x^2+2)^2$ using the chain rule is $f'(x) = 2(x^2+2)(2x) = 4x(x^2+2)$.

   The gradient at $x=0$ is $m = f'(0) = 4(0)(0^2+2) = 0$.

   Since the gradient is zero, it is proven that the tangent line is parallel to the $x$-axis.

   Step 3: Prove its distance is $4 \text{ units}$ from the origin.

   The equation of the tangent line at point $(0,4)$ with gradient $m=0$ is:

   $$y - 4 = 0(x - 0) \implies y = 4$$

   The line $y=4$ is a horizontal line. The distance from any point on this line to the $x$-axis (the line $y=0$) is $4 \text{ units}$. Since the origin $(0,0)$ lies on the $x$-axis, the distance from this tangent line to the origin is also $4 \text{ units}$. Proven.

5. Solution:

   Step 1: Determine the gradient from the angle.

   The relationship between the gradient ($m$) and the angle ($\theta$) a line makes with the positive $x$-axis is given by $m = \tan(\theta)$.

   $$m = \tan(45^\circ) = 1$$

   The required tangent line has slope $1$.

   Step 2: Find the coordinates of the point of tangency.

   The gradient is also the first derivative of the curve $f(x) = 2x^2 - 7x + 1$.

   $$f'(x) = 4x - 7$$

   We set it equal to the gradient we found:

   $$4x - 7 = 1$$

   $$4x = 8 \implies x = 2$$

   Now, find the $y$-value by plugging $x=2$ into the curve's equation:

   $$\begin{aligned} y &= 2(2)^2 - 7(2) + 1 \\ &= 8 - 14 + 1 \\ &= -5 \end{aligned}$$

   The point of tangency is $(2, -5)$.

   Step 3: Determine the equation of the tangent line.

   Using the point $(2, -5)$ and gradient $m=1$:

   $$y - (-5) = 1(x - 2)$$

   $$y + 5 = x - 2$$

   $$y = x - 7$$