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Find the line passing through the point (-3,0) and tangent to the curve y = sqrt(x-1) at some point. Note The point (-3,0) is not on the given curve.
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Note that this type of function is going to have only one tangential point with the line. This would not be true with a function like a parabola. The hard part conceptualizing this using variables instead of constants. So lets set up a couple of algebraic equations. Our slope of the line will equal y/(x+3). This will be the same slope as the derivative of y=sqrt(x-1). You can set the the derivative and your slope equations equal. ie y'=y/(x+3) or y'=[sqrt(x-1)]/(x+3). Once you find x you can find the slope and write your equation. good luck
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