# Find the equation of a tangent line to y=f(x)

## Homework Statement

Find the equation of the tangent line to the curve at the given point.
$$y = \sqrt{2x+1} , (4, 3)$$

## Homework Equations

a = 4
$$M =\lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$

or

$$Mpq = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

## The Attempt at a Solution

$$M =\lim_{x \to 4} \frac{f(x) - f(4)}{x-a}$$

$$M =\lim_{x \to 4} \frac{\sqrt{2x+1} - \sqrt{2 * 4 + 1}}{x-4}$$

$$M =\lim_{x \to 4} \frac{\sqrt{2x+1} - 3}{x-4}$$

$$M =\lim_{x \to 4} \frac{\sqrt{2x+1} - 3}{x-4} * \frac{\sqrt{2x+1} + 3}{\sqrt{2x+1} + 3}$$

$$M =\lim_{x \to 4} \frac{2x - 3}{x-4(\sqrt{2x+1} + 3)}$$

At this point I get stuck... I'm not sure what I can do from here. did I go wrong some where?

On the last step are you sure that you multiplied correctly? :P (check the numerator again)

Do you see it?

It should be 2x-8 since $$(\sqrt{2x+1}-3) * (\sqrt{2x+1}+3) = 2x+1 -9 = 2x-8$$

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From that point you can divide by (x-4) quite easily resulting in a function for which you can directly substitute x=4 since the result is not in an 'indeterminate form' of $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$

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BAH! thanks :).