Tangent Line Calculator

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Tangent line is a line that touches the given curve at some point (x_{o},y_{o}) such that slope of the curve will be equal to the slope of the line.If the curve touches the curve at a point (x_{o},y_{o}). The slope for the tangent line is given by (dy/dx) (x = x_{o}).

**Figure showing the tangent line**

Tangent line calculator is a amazing online tool that finds the equation of tangent line for any equation you enter and even gives you out the graph instantly within the fraction of seconds.

A function is given below as default input for this calculator. The given function is differentiated and slope point is calculated to get the equation of tangent line. You can see graph plotted when you click on "Graph".

A function is given below as default input for this calculator. The given function is differentiated and slope point is calculated to get the equation of tangent line. You can see graph plotted when you click on "Graph".

Read the given problem and observe whether the given equation y = f(x) and note down the initial points (x_{0},y_{0}).

Differentiate the given equation y = f (x) to get f'(x).

Plug in the value of xo in f'(x) to get f'(x) is the slope m.

Use the slope intercept form y - yo = m(x - xo). Plug in the values of (xo, yo) to get the equation of line.

Find the equation of the line tangent to the curve f(x) = 2x

^{2}- x + 3 at x = 1?**Step 1 :**The given function is f(x) = 2x

^{2}- x + 3, initial point x_{o}= 1 and y_{o}= 2(1)^{2}- 1 + 3 = 4**Step 2 :**Differentiate the given function f'(x) = $\frac{dy}{dx}$ = $\frac{d(2x^2 - x + 3)}{dx}$ = 4x - 1

f'(1) = 4(1) - 1 = 3**Step 3 :**The slope point form is y - y

_{o}= m (x - x_{o})

y - 4 = 3 (x - 1)

Simplifying we get

y - 4 = 3x -1

y = 3x + 3**Step 4 :**The equation of tangent line is y = 3x + 3. The graph for this line is

**Answer :**y = 3x + 3

Find the equation of the line tangent to the curve f(x) = x

^{3}at x = 2?**Step 1 :**The given function is f(x) = x

^{3}, initial point x_{o }= 2 and y_{o}= 2^{3}= 8**Step 2 :**Differentiate the given function f'(x) = $\frac{dy}{dx}$ = $\frac{d(x^3))}{dx}$ = 3x

^{2}

f'(2) = 3(2)^{2}= 12**Step 3 :**The slope point form is y - y

_{o}= m (x - x_{o})

y - 8 = 12 (x - 2)

Simplifying we get y - 8 = 12x - 24

y = 12x - 16.**Step 4 :**The equation of tangent line is y = 12x - 16. The graph for this line is

**Answer :**y = 12x - 16