The equation for the tangent line to a curve can be found by finding the slope of the tangent line and using the coordinates of the given point. In turn, we can determine the slope of the tangent line by differentiating the function and using the value of *x* from the given point in that derivative.

Here, we will learn how to find the equation of the tangent line to a curve step by step. We will learn about the process that we can use, and we will apply it to solve some practice problems.

## Process to find the equation for the tangent line to a curve

The equation of the tangent line to a curve can be found using the form $latex y=mx+b$, where *m* is the slope of the line and *b* is the y-intercept.

Therefore, if we want to find the equation of the tangent line to a curve at the point $latex (x_{1},~y_{1})$, we can follow these steps:

**Step 1:** Find the derivative of the function that represents the curve.

**Step 2:** Use the derivative of the function to find the slope of the tangent line at the point $latex (x_{1},~y_{1})$. For this, we use the

*x*-coordinate of the point in the derivative of the function. That is, we have $latex m=f'(x_{1})$.

**Step 3:** Use the equation $latex y=mx+b$ with the value of the slope found in step 2 and substitute the

*x*and

*y*coordinates of the given point to find the value of

*b*. That is, we have $latex y_{1}=mx_{1}=b$.

**Step 4:** Substitute the values of

*m*and

*b*into the equation $latex y=mx+b$.

Look at the following examples to learn how to apply these steps to find the equation of the tangent line to a function.

## Examples of the equation for the tangent line to a curve

In the following examples, we apply the steps seen above to find the equation of the tangent line to each of the following functions.

**EXAMPLE 1**

Find the equation of the tangent line to the curve $latex f(x)=x^2$ at the point P=(2, 4).

##### Solution

**Step 1:** The derivative of the function is:

$latex f(x)=x^2$

$latex f'(x)=2x$

**Step 2:** We find the slope of the tangent line at the point (2, 4) by evaluating $latex f'(2)$. Then, we have:

$latex m=f'(2)=2(2)$

$latex m=4$

**Step 3:** Now, we have the equation $latex y=4x+b$. To find the value of

*b*, we use the point (2, 4) in the equation:

$latex y=4x+b$

$latex 4=4(2)+b$

$latex b=-4$

**Step 4:** The equation of the tangent line at the point (2, 4) is $latex y=4x-4$.

**EXAMPLE **2

**EXAMPLE**What is the equation of the tangent to $latex f(x)=3x^2-3x$ at the point (1, 3)?

##### Solution

**Step 1:** The derivative of the function is:

$latex f(x)=3x^2-3x$

$latex f'(x)=6x-3$

**Step 2:** The slope of the tangent line at the point (1, 3) is found by evaluating $latex f'(1)$. Therefore, we have:

$latex m=f'(1)=6(1)-3$

$latex m=3$

**Step 3:** With the slope from step 2 we form the equation $latex y=3x+b$. Then, we find the value of

*b*, using the point (1, 3) in the equation:

$latex y=3x+b$

$latex 3=3(1)+b$

$latex b=0$

**Step 4:** The equation of the tangent line at the point (1, 3) is $latex y=3x$.

**EXAMPLE **3

**EXAMPLE**Find the equation of the tangent line to $latex f(x)=x^3+\frac{4}{x}$ at the point (2, 5).

##### Solution

**Step 1:** We write the radical expression as a numerical exponent and find its derivative:

$latex f(x)=x^3+4x^{-1}$

$latex f'(x)=3x^2-4x^{-2}$

$latex f'(x)=3x^2-\frac{4}{x^2}$

**Step 2:** We evaluate $latex f'(2)$ to find the slope of the tangent line at (2, 5). Thus, we have:

$latex m=f'(2)=3(2)^2-\frac{4}{2^2}$

$latex =12-1$

$latex m=11$

**Step 3:** With the slope from step 2, we have the equation $latex y=11x+b$. To find the value of

*b*, we use the point (2, 5) in the equation:

$latex y=11x+b$

$latex 5=11(2)+b$

$latex b=-17$

**Step 4:** The equation of the tangent line at the point (2, 5) is $latex y=3x-17$.

**EXAMPLE **4

**EXAMPLE**Find the tangent line to $latex f(x) = -x^{-2}+\sqrt{x}$ at the point (1, 3).

##### Solution

**Step 1:** We write the square root as a numerical exponent and find the derivative of the function:

$latex f(x)=-x^{-2}+x^{\frac{1}{2}}$

$$f'(x)=2x^{-3}+\frac{1}{2}x^{-\frac{1}{2}}$$

$$f'(x)=\frac{2}{x^3}+\frac{1}{2\sqrt{x}}$$

**Step 2:** We evaluate $latex f'(1)$ to find the slope of the tangent line at (1, 3). Therefore, we have:

$$m=f'(1)=\frac{2}{(1)^3}+\frac{1}{2\sqrt{1}}$$

$latex =2+\frac{1}{2}$

$latex m=\frac{5}{2}$

**Step 3:** Now, we have the equation $latex y=\frac{5}{2}x+b$. To find the value of

*b*, we use the point (1, 3) in the equation:

$$y=\frac{5}{2}x+b$$

$$3=\frac{5}{2}(1)+b$$

$latex b=\frac{1}{2}$

**Step 4:** The equation of the tangent line at the point (1, 3) is $latex y=\frac{5}{2}x+\frac{1}{2}$.

**EXAMPLE **5

**EXAMPLE**What is the equation of the tangent line to $latex f(x)=\sin(x)-\cos(x)$ at the point (0, 1)?

##### Solution

**Step 1:** The derivative of the function is:

$latex f(x)=\sin(x)-\cos(x)$

$latex f'(x)=\cos(x)+\sin(x)$

**Step 2:** We find the slope of the tangent line at the point (0, 1) by evaluating $latex f'(0)$. Then, we have:

$latex m=f'(0)=\cos(0)+\sin(0)$

$latex m=1+0$

$latex m=1$

**Step 3:** Using the slope $latex m=1$, we have the equation $latex y=x+b$. Now, we find the value of

*b*, using the point (0, 1) in the equation:

$latex y=x+b$

$latex 1=0+b$

$latex b=1$

**Step 4:** The equation of the tangent line at the point (0, 1) is $latex y=x+1$.

**EXAMPLE **6

**EXAMPLE**Find the tangent line to $latex f(x)=x^2-3x+1$ at the point where the curve intersects the *y*-axis.

##### Solution

**Step 1:** We start by finding the derivative of the function:

$latex f(x)=x^2-3x+1$

$latex f'(x)=2x-3$

**Step 2:** The problem doesn’t give us a point directly, but tells us to find the tangent at the point where the curve intersects the

*y*-axis. Therefore, we have to find the coordinates of that point.

When the curve intersects the *y*-axis, we know that the *x*-coordinates must be 0. Then, we have:

$latex y=x^2-3x+1$

$latex y=0^2-3(0)+1$

$latex y=1$

Thus, the point is (0, 1). We find the slope of the tangent line at the point (0, 1) by evaluating $latex f'(0)$. Then, we have:

$latex m=f'(0)=2(0)-3$

$latex m=-3$

**Step 3:** We have found the equation $latex y=-3x+b$. To find the value of

*b*, we use the point (0, 1) in the equation:

$latex y=-3x+b$

$latex 1=-3(0)+b$

$latex b=1$

**Step 4:** The equation of the tangent line at the point (0, 1) is $latex y=-3x+1$.

## Equation of the tangent line to a curve – Practice problems

Solve the following practice problems by using everything you have learned about the equation for the tangent line to a curve.

## See also

Interested in learning more about equations of tangent and normal lines to functions? You can take a look at these pages:

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