# Find all tangent lines to the graph of #y = x^3# that pass through the point. #P(2, 4)#?

Setting this up yields:

Now solving:

Find the slopes of the tangent lines by finding the value of the derivative at each of these points:

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To find the tangent lines to the graph of y = x^3 that pass through the point P(2, 4), we can use the point-slope form of a line.

First, we differentiate the equation y = x^3 to find the slope of the tangent line at any given point on the graph. The derivative of y = x^3 is dy/dx = 3x^2.

Next, we substitute the x-coordinate of the point P(2, 4) into the derivative to find the slope at that point. dy/dx = 3(2)^2 = 12.

Now, we have the slope of the tangent line at the point P(2, 4). Using the point-slope form of a line, which is y - y1 = m(x - x1), we substitute the values of the point and slope into the equation.

y - 4 = 12(x - 2)

Simplifying the equation, we get:

y - 4 = 12x - 24

Rearranging the equation to the slope-intercept form, we have:

y = 12x - 20

Therefore, the equation of the tangent line to the graph of y = x^3 that passes through the point P(2, 4) is y = 12x - 20.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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