Equation of a Tangent Line: Problems and Solutions

Equation of a Tangent Line or Normal Line: Problems and Solutions

Are you working to find the equation of a tangent line (or normal line) in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

CALCULUS SUMMARY: Tangent & Normal Lines

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These problems will always specify that you find the tangent or normal (= perpendicular) line at a particular point of a function. We’ll call that point $(x_0, y_0)$.

To answer these questions, you will almost always use the Point-Slope form of a line. Recall that if a line has slope m and contains the point $(x_0, y_0)$, then you can write its equation as:

Point-Slope form of a line: $$\bbox[yellow,5px]{y – y_0 = m(x – x_0)}$$

The problem statement typically specifies the point $(x_0, y_0)$, and so really these problems come down to determining the slope m of the line — which we’ll address below.

You will use that equation again and again; memorize it if you don’t know it already.

(It’s just a variant on the definition of slope: $m = \dfrac{y – y_0}{x – x_0}.)$

I. Tangent Line to a Curve

Very frequently in beginning Calculus you will be asked to find an equation for the line tangent to a curve at a particular point. We’re calling that point $(x_0, y_0)$.

To find the line’s equation, you just need to remember that the tangent line to the curve has slope equal to the derivative of the function evaluated at the point of interest:

If those equations look abstract to you, don’t worry. As soon as you work a few problems, the process will make sense — we promise.

II. Normal Line to a Curve

Sometimes instead a question will ask you instead to find the line normal to a curve. That’s the same thing as asking for the line that is perpendicular to the curve.

You will again use the Point-Slope form of a line. But now to compute the slope of the line, recall that the slopes of perpendicular lines are the negative reciprocals of each other ($m_2 = -\dfrac{1}{m_1}$). We want the slope of the line that is perpendicular to the curve at a point, and hence that is perpendicular to the tangent line to the curve at that point:

\[\bbox[yellow,5px]{ \begin{align*} m_\text{normal line} &= \frac{-1}{m_\text{tangent line}}\\[12px]
&= \frac{-1}{f'(x_0)} \end{align*}}\] Hence we can write the equation for the normal line at $(x_0, y_0)$ as

We recommend not trying to memorize all of the formulas above. Instead, remember the Point-Slope form of a line, and then use what you know about the derivative telling you the slope of the tangent line at a given point. The problems below illustrate.

Problem 1 illustrates the process of putting together different pieces of information to find the equation of a tangent line.

Problem 2 requires that you find the pieces of information before you can put them together.

Tangent/Normal Line Problem #1

For a particular function $g$, we know $g'(5) = 2$ and $g(5) = -3$. Write an equation for the line tangent to $g$ at $x = 5$.

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Since $g'(5) = 2$, we know that the slope of the tangent line at $x = 5$ is $m=2$. Furthermore, the tangent line contains the point (5, -3), since it passes through (grazes) that point on the curve.

Then using the point-slope form of a line that contains the point $(x_o, y_o)$ we have \begin{align*} y – y_o &= m(x – x_o) \\ y – (-3) &= 2(x-5) \\ y + 3 &= 2(x-5) \quad \cmark \end{align*} We can stop there, since we have found a valid equation for the line. OR, we can continue and put the equation into slope-intercept form: \begin{align*} y+3 &= 2x – 10 \\ y &= 2x – 10 -3 \\ y &= 2x – 13 \quad \cmark \end{align*} Either answer is valid and correct.

Tangent/Normal Line Problem #2

Consider the curve given by $y = f(x) = x^3 – x + 5$. (a) Find the equation to the line tangent to the curve at the point (1, 5). (b) Find the equation of the line normal (perpendicular) to the curve at the point (1, 5).

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Solution (a)Solution (b)

To write the equation of a line, we need its slope $m$ and a point $(x_o, y_o)$ on it. We already know a point, since the line intersects (grazes) the curve at (1, 5).

Hence we just need its slope m, which is is the same as the slope of the curve at that point $(x=1).$ And that slope equals the function’s derivative at that point: \begin{align*} f(x) &= x^3 – x + 5\\[8px]
f'(x) &= 3x^2 -1 \\[8px]
f'(1) &= 3(1) – 1 = 2 \end{align*} Hence the tangent line has slope $m_\text{tangent} = f'(1) = 2.$

Then we can write the equation of this tangent line, using the point-slope form of a line that has slope $m_\text{tangent} = 2$ and contains the point $(1, 5)$: \begin{align*} y – y_o &= m_\text{tangent}(x – x_o) \\[8px]
y – 5 &= (2)(x – 1) \quad \cmark \end{align*} We can stop there, since we have found a valid equation for the line. OR, we can continue and put the equation into slope-intercept form: \begin{align*} y – 5 &= 2x -2 \\[8px]
y &= 2x – 2 + 5 \\[8px]
y&= 2x +3 \quad \cmark \end{align*} Either answer is valid and correct.

Recall that line 2 is normal (perpendicular) to line 1 if their slopes are negative reciprocals: $$m_2 = -\frac{1}{m_1}$$ From part (a) we know that the line tangent to the curve at the point (1,5) has slope $m_\text{tangent} = 2$, and so our perpendicular line at that point has slope \begin{align*} m_\text{normal} &= -\frac{1}{m_\text{tangent}} \\[8px]
&= -\frac{1}{2} \end{align*} Hence the normal line has slope $m_\text{normal} = -\dfrac{1}{2}.$

This line passes through the curve at point (1, 5) and has slope $m_\text{normal} = -\dfrac{1}{2}$, and so its equation is given by \begin{align*} y – y_o &= m_\text{normal}(x – x_o) \\[8px]
y-5 &= -\dfrac{1}{2} (x -1) \quad \cmark \end{align*} We can stop there, since we have found a valid equation for the line. OR, we can continue and put the equation into slope-intercept form: \begin{align*} y – 5 &= -\dfrac{1}{2}x + \dfrac{1}{2} \\[8px]
y &= -\dfrac{1}{2}x + \dfrac{1}{2} + 5 \\[8px]
y &= -\dfrac{1}{2} x + \frac{11}{2} \quad \cmark \end{align*} Either answer is valid and correct.

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Following the approach outlined in the SUMMARY box above:

1. The slope of the tangent line at (1,1) equals the derivative at that point. Since $y=f(x)= x^3,$ the derivative is $y’=f'(x) =3x^2.$ Hence the slope of the tangent line at $x_0 = 1$ is $m_{tangent} = f'(1) = 3.$

2. Then we can write the equation of this tangent line, using the point-slope form of the line that has slope $m_{tangent} = 3$ and contains the point (1,1): $$ y – 1 = 3(x-1)$$

Does that provide the solution you were after? We hope so!

Check out our free materials: Full detailed and clear solutions to typical problems, and concise problem-solving strategies.

Y=x³at the point 1,1 give the solution

Following the approach outlined in the SUMMARY box above:

1. The slope of the tangent line at (1,1) equals the derivative at that point. Since $y=f(x)= x^3,$ the derivative is $y’=f'(x) =3x^2.$ Hence the slope of the tangent line at $x_0 = 1$ is $m_{tangent} = f'(1) = 3.$

2. Then we can write the equation of this tangent line, using the point-slope form of the line that has slope $m_{tangent} = 3$ and contains the point (1,1):

$$ y – 1 = 3(x-1)$$

Does that provide the solution you were after? We hope so!