Equation of tangent to a curve.

**darksoulzero** · July 12th 2010, 07:34 PM

Hi, I have a question that asks: Write an equation of the tangent to each curve at the given point. $y = \sqrt{3x^3}$

I've been trying to do the question, but I end up with decimals in the equation for the tangent. The answer is supposed to be $9x -2y - 9 = 0$ . Thanks in advance.

EDIT: Sorry about that, the given point is $P(3,9)$

**pickslides** · July 12th 2010, 07:47 PM

Start by finding $\frac{dy}{dx}$ at $x=3$

**darksoulzero** · July 12th 2010, 08:04 PM

Originally Posted by pickslides

Start by finding $\frac{dy}{dx}$ at $x=3$

Thanks, but I think that something's wrong with my math. here's what I got:

$\frac{dy}{dx} = \frac{9}{2} \sqrt{x}.$

$f'(3) = \frac{9}{2} \sqrt{3}$

Now I sub. f'(3) into the following equation:

$y - y_{1} = m(x - x_{1})$

$y - 9 = \frac{9}{2} \sqrt{3}x -\frac{9}{2} \sqrt{3} (3)$

$y - 9 - \frac{9}{2} \sqrt{3}x + \frac{9}{2} \sqrt{3}(3) = 0$

From here, I'm stumped as to how to get to $9x -2y - 9 = 0$ .

**darksoulzero** · July 12th 2010, 08:23 PM

Nevermind, I got the answer.

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