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Tangent and velocity

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The parabola y = (x^2) + 3 has two tangents which pass through the point (0, -2). One is tangent to the to the parabola at (A, A^2 + 3) and the other at (-A, A^2 + 3). Find (the positive number) ?

If a ball is thrown vertically upward from the roof of 64ft foot building with a velocity of 96 ft/sec, its height after t seconds is s(t) = 64 + 96t - 16t^2. I found it's max height to be 208 ft. What is the velocity of the ball when it hits the ground (height0)?

Use the derivative to find this equation of the tangent line to the curve y = 4x + 3 * square root of x at the point (4, 22.000000). The equation of this tangent line can be written in the form y = mx + b?

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Solution Summary

The three problems in this solution cover the following: finding the point where a given tangent passes through a parabola, the velocity of a ball when it hits the ground, and the equation of a tangent line.

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Let the equation of the parabola be f(x) = x^2 + 3. The slope of the tangent at point (x1,y1) is f'(x1).
Therefore, the slope of the tangent at (A,A^2+3) = f'(A) = 2A ......1
Since the tangent passes through (0,-2) and (A,A^2+3), the slope can be calculated as = {A^2+3-(-2)}/{A-0} = {A^2+5}/A ...

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