# `f(x) = x^3, (2, 8)` Find an equation of the tangent line to the graph of f at the given point.

Given: `f(x)=x^3`   and the coordinate (2, 8)` `

To find the slope take the derivative of the function and evaluate the derivative at x=2

`f'(x)=3x^2`

`f'(2)=3(2)^2`

`f'(2)=12`

The slope is 12. Use the slope of 12 and the coordinate (2, 8) to find the equation of the line tangent to the graph at (2, 8). The point slope form of an equation is. The slope is 12 and (x1, y1) is the coordinate (2, 8).

`y-y1=m(x-x1)`

`y-8=12(x-2)`

`y-8=12x-24`

`y=12x-24+8`

`y=12x-16`

Therefore the equation is: ``

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The given function is:-

f(x) = (x^3)

differentiating both sides w.r.t 'x' we get

f'(x) = 3*(x^2)

Now, slope of the tangent at the point (2,8) = f'(2) = 3*(2^2) = 12

Thus, equation of the tangent at the point (2,8) and having slope = 12 is :-

y - 8 = (12)*(x - 2)

or, y - 8 = 12x - 24

or, y - 12x + 16 = 0 is the equation of the tangent to the  given curve at (2,8)

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