# Tangent Line Calculator

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In Geometry, a straight line which touches a plane curve at a given point is called the tangent line to the curve or just tangent. Tangent line always just touches the surface of the curve at a point. The point at which it touches the curve is called the point of tangency.

In Calculus, the tangent line is a straight line to a given curve  ${y = f(x)}$  at a point  $x_{0}$ , which passes through the point  $(x_{0}, f(x_{0}))$ on the curve and its slope is given by  $f'(x_{0})$, where $f'(x)$ represents the derivative of the function $f(x)$.

## How to Find Tangent Line

Steps to find the slope of a tangent and equation of tangent line :
Step 1: Identify the given equation and the given point at which the tangent equation has to be defined
Step 2: Write the equation in the form of   ${y = f(x)}$.
Step 3: Differentiate the equation  ${y = f(x)}$ to get $\frac{dy}{dx}$ $= f'(x)$.

Step 4:
Plug the given point $x_{0}$ in the derivative i.e., $\frac{dy}{dx}$ $= f'(x_{0})$.
Step 5: The slope of the tangent, $m = f'(x_{0})$
Step 6: Use the point slope form  $y - y_{0} = m(x - x_{0})$ and plug the values of $x_{0}, y_{0}$
Step 7: The obtained equation is the equation of line.

The slope of a tangent line parallel to $x$ axis is zero and the slope of the tangent line which is parallel to $y$ – axis is not defined.

## Tangent Line Examples

### Solved Examples

Question 1: What is the slope of the tangent line to the curve $y = 5x^{2} - 3x + 2$ at the point $(1, 1)$?
Solution:

Given $y = 5x^{2} - 3x + 2$
Slope of the tangent to the given curve at any point $(x_{0}, y_{0})$ is given by its derivative at $x_{0}$ i.e., $\frac{dy}{dx}$$_{x = x_{0}}. Differentiating with respect to x on both sides of the equation, \frac{dy}{dx} =$$\frac{d}{dx}$$(5x^{2} - 3x + 2) Use Power rule of derivatives \frac{d}{dx} x^{n} = n x^{n - 1} \frac{dy}{dx} = 10x - 3 Plug x = 1 to get the slope at (1,1) \frac{dy}{dx}$$ _{x = 1}$ $= 10(1) - 3 = 7$
So, slope of the tangent to the curve $y = 5x^{2} - 3x + 2$ at $(1, 1)$ is $7$.

Question 2: What is the x coordinate of the point at which the curve $y = 2x^{3} - 7$ has a slope $12$?
Solution:

Given $y = 2x^{3} - 7$
Slope of the tangent to the given curve at any point $(x_{0}, y_{0})$ is given by its derivative at $x_{0}$ i.e., $\frac{dy}{dx}$$_{x = x_{0}} Differentiating with respect to x on both sides of the equation, \frac{dy}{dx}$$ = $$\frac{d}{dx}$$ (2x^{3} - 7)$
Use Power rule of derivatives $\frac{d}{dx}$ $x^{n} = n x^{n - 1}$
$\frac{dy}{dx}$ $= 6x$
Given the slope at a point $x = x_{0}$ is $12$,
$\frac{dy}{dx}$$_{x = x_{0}} = 6x_{0} 12 = 6x_{0} Dividing both sides by 6, x_{0} =$$\frac{12}{6}$$= 2 Thus, the x coordinate of the point at which the curve y = 2x^{3} - 7 has a slope 12 is 2. Question 3: Find the equation of the tangent line to the curve y = 2x^{2} - 7x + 4 at the point (2, 4). Solution: Given y = 2x^{2} - 7x + 4 Slope of the tangent to the given curve at any point (x_{0}, y_{0}) is given by its derivative at x_{0} i.e., \frac{dy}{dx}$$ _{x = x_{0}}$

Differentiating with respect to $x$ on both sides of the equation,
$\frac{dy}{dx}$$=$$\frac{d}{dx}$$(2x^{2} - 7x + 4) Use Power rule of derivatives \frac{d}{dx} x^{n} = n x^{n - 1} \frac{dy}{dx} = 4x - 7 Plug x = 1 to get the slope at (2, 4) \frac{dy}{dx}$$ _{x = 1}$ $= 4(2) - 7 = 1$
So, slope of the tangent to the curve $y = 2x^{2} - 7x + 4$ at $(2, 4)$ is $1$.
Use the point slope form $y - y_{0} = m(x - x_{0})$ and plug the values $x_{0} = 2, y_{0} = 4$ and $m = 1$.
$y - 4 = 1(x - 2)$
$y - 4 = x - 2$
Adding $4$ to both sides of the equation,
$y = x - 2 + 4 = x + 2$
Thus, the required tangent line equation is $y = x + 2$