# How do you find the equation of the tangent line to the curve #y=1+x^3# that is parallel to the line #12x-y=1#?

Rearranging the eq of the line to gradient-intercept form,

The first derivative of the curve would be the gradient of the curve, as well as the gradient of tangents to the curve at different points all along the curve.

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To find the equation of the tangent line to the curve y=1+x^3 that is parallel to the line 12x-y=1, we need to find the derivative of the curve and then use it to determine the slope of the tangent line.

The derivative of y=1+x^3 is dy/dx = 3x^2.

Since the tangent line is parallel to the given line, it will have the same slope. The slope of the given line can be determined by rearranging it into the slope-intercept form y = mx + b, where m represents the slope.

12x - y = 1 can be rewritten as y = 12x - 1.

Comparing this equation with y = mx + b, we can see that the slope of the given line is 12.

Since the slope of the tangent line is also 12, we can set dy/dx = 12 and solve for x.

3x^2 = 12

Dividing both sides by 3, we get x^2 = 4.

Taking the square root of both sides, we find x = ±2.

Now, we substitute the x-values into the original equation y = 1+x^3 to find the corresponding y-values.

For x = 2, y = 1 + (2)^3 = 9.

For x = -2, y = 1 + (-2)^3 = -7.

So, we have two points on the curve: (2, 9) and (-2, -7).

Using the point-slope form of a line, y - y1 = m(x - x1), we can choose one of the points and the slope (m = 12) to find the equation of the tangent line.

Using the point (2, 9), we have y - 9 = 12(x - 2).

Simplifying this equation, we get y - 9 = 12x - 24.

Rearranging it into the slope-intercept form, we have y = 12x - 15.

Therefore, the equation of the tangent line to the curve y=1+x^3 that is parallel to the line 12x-y=1 is y = 12x - 15.

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