To find the equation of the tangent line to the curve y=x^3 at the point (2,8), we need to find the slope of the tangent line at that point.
The slope of a tangent line to a curve at a given point can be found by taking the derivative of the function representing the curve and evaluating it at the x-coordinate of the given point.
The derivative of y=x^3 is given by:
dy/dx = 3x^2.
To find the slope at x=2, we evaluate the derivative at x=2:
dy/dx = 3(2)^2 = 12.
Therefore, the slope of the tangent line at the point (2,8) is 12.
Now that we have the slope, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1),
where (x1, y1) is a point on the line and m is the slope.
Using the point (2,8) and the slope 12, we can write the equation of the tangent line as:
y - 8 = 12(x - 2).
Expanding and rearranging the equation, we get:
y - 8 = 12x - 24,
y = 12x - 24 + 8,
y = 12x - 16.
Therefore, the equation of the tangent line to the curve y=x^3 at the point (2,8) is y = 12x - 16.
Find the equation of the tangent to the curve y=x3 at the point (2,8)
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