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Tangent Line Calculator

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Tangent line calculator finds the equation of a tangent line of an equation. Tangent line is a line which touches the curve at one single point. Below figure shows tangent line to a curve at a point ($x_0$, $y_0$). The slope of the tangent line is the derivative of the function f(x).

Tangent Line
 

Steps

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Step 1 : Observe the given equation and point and rewrite the equation in y = f(x) form.

Step 2 : Differentiate the given equation y = f(x) and get $\frac{dy}{dx}$ $= f'(x)$.

Step 3 : Substitute $x_0$ into f'(x) to get the slope of the line m.

Step 4 : Substitute the values of ($x_0$, $y_0$) into the slope-intercept form $y - y_0 = m(x - x_0)$ to get the equation of line.

Problems

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Below are the problems based on Tangent Line Calculator.

Solved Examples

Question 1: Find the equation of the tangent line of the following function with graph : $f(x) = x^3 + 2x$ at x = 1 ?
Solution:
 
Step 1 : Given :
         $f(x) = x^3 + 2x$, 
         $x_0$ = 1
         $y_o$ = $(1)^3 + 2(1)$ = 3
   
Step 2 : $\frac{dy}{dx}$ = $f'(x)$ = $\frac{d}{dx}$$(x^3 + 2x)$ = $3x^3 + 2$
       
Step 3 : f(1) = 3(1)^3 + 2 = 5 
           Therefore, slope of the tangent lime, m = 5
         
Step 4 : By simplifying $x_0$ and $y_0$ in the slope-intercept form,
            $y - y_0 = m(x - x_0)$
            $y - 3 = 5(x - 1)$
            $y = 5x - 2$
         
Therefore, the equation of line is $y = 5x - 2$. 

The graph for the following equation is shown below.
Tangent Line Graph

 

Question 2: Find the equation of the tangent line of the following function with graph : $f(x) = 4x^2 - 4x + 1$ at x = 0 ?
Solution:
 
Step 1 : Given :
         $f(x) = 4x^2 - 4x + 1$, 
         $x_0$ = 0
         $y_o$ = $4(0)^2 - 4(0) + 1$ = 1 
   
Step 2 : $\frac{dy}{dx}$ = $f'(x)$ = $\frac{d}{dx}$$(4x^2 - 4x + 1)$ = $8x - 4$
       
Step 3 : f(0) = 8(0) - 4 = - 4
           Therefore, slope of the tangent lime, m = - 4
         
Step 4 : By simplifying $x_0$ and $y_0$ in the slope-intercept form,
            $y - y_0 = m(x - x_0)$
            $y - 1 = - 4(x - 0)$
            $y = - 4x + 1$
         
Therefore, the equation of line is $y = - 4x + 1$. 

The graph for the following equation is shown below.
Graph of a Tangent Line