How do you find the equation of the tangent line to the graph of #f(x)=x^3# at point (2,8)?

Question

Christopher Agresta · Accepted Answer

# y = 12 x - 16 #

The gradient of the tangent at any particular point is given by the derivative.

# f(x) = x^3 #
# :. f'(x) = 3x^2 #

At # (2,8), x=2 => f'(2) = (3)2^2 = 12 #

So the tangent passes through #(2,8)# and has gradient #m=12#
So, Using # y - y_1 = m(x - x_1 ) #, the tangent equation is

# y - 8 = 12( x - 2 ) #
# :. y - 8 = 12 x - 24 #
# :. y = 12 x - 16 #

Cameron Alexander · Answer

The method you use to find the slope of the tangent line depends on your current studies.

If you have learned the power rule for derivatives, use that. If you are using the definition (or an equivalent), then the details depend on the definition (or equivalent) you use. Two likely definitions are The slope of the line tangent to the graph of function #f# at the point #(a,f(a))# is given by: #lim_(hrarr0)(f(a+h)-f(a))/h# #" "# OR #" "# #lim_(trarra)(f(t)-f(a))/(t-a)# In this question, we have #f(x) = x^3# and #a = 2# Using the first version #lim_(hrarr0)(f(2+h)-f(2))/h = lim_(hrarr0)((2+h)^3- (2)^3)/h # # = lim_(hrarr0)(8+12h+6h^2+h^3- 8)/h # # = lim_(hrarr0)(12h+6h^2+h^3)/h # # = lim_(hrarr0)(cancel(h)(12+6h+h^2))/cancel(h) # # = 12 + 6(0)+(0)^2 = 12# Using the second version #lim_(trarr2)(f(t)-f(2))/(t-2) = lim_(trarr2)((t)^3-(2)^3)/(t-2)# # = lim_(trarr2)(t^3- 8)/(t-2)# # = lim_(trarr2)(cancel((t-2))(t^2+2t+4))/cancel(t-2)# # = (2)^2+2(2)+4 = 4+4+4=12# Now that we have the slope, we can do the algebra to find an equation for the line through #(2,8)# with slope #m=12# #y=12x-16#

Evelyn Agard · Answer

undefined To find the equation of the tangent line to the graph of f(x)=x^3 at point (2,8), we need to find the slope of the tangent line at that point. The slope of the tangent line is equal to the derivative of the function at that point.

Taking the derivative of f(x)=x^3, we get f'(x) = 3x^2.

To find the slope at x=2, we substitute x=2 into the derivative: f'(2) = 3(2)^2 = 12.

So, the slope of the tangent line at point (2,8) is 12.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values:

y - 8 = 12(x - 2).

Simplifying the equation, we get y - 8 = 12x - 24.

Rearranging the equation, we have y = 12x - 16.

Therefore, the equation of the tangent line to the graph of f(x)=x^3 at point (2,8) is y = 12x - 16.