finding eq. of tan to the slope f(x) given the slope

**tecknics** · September 16th 2009, 06:43 PM

I know how to do this problem but I just totally had a brain fade and cant seem to figure it out, can someone lead me into the right direction here?

**DeMath** · September 16th 2009, 07:58 PM

Originally Posted by tecknics

I know how to do this problem but I just totally had a brain fade and cant seem to figure it out, can someone lead me into the right direction here?

$y = \sqrt {x + 9}.$

An equation of the tangent
$\boxed{y = {{y'}_0}\left( {x - {x_0}} \right) + {y_0}}$

By the condition of your task, you have

${{y'}_0} = {\left( {\sqrt {{x_0} + 9} } \right)^\prime } = \frac{1}<br /> {4} \, \Rightarrow \, \frac{1}{{2\sqrt {{x_0} + 9} }} = \frac{1}{4} \Leftrightarrow \sqrt {{x_0} + 9} = 2 \Rightarrow {x_0} = - 5.$

${y_0} = \sqrt { - 5 + 9} = \sqrt 4 = 2.$

Finally, you get this equation of the tangent

$y = \frac{1}{4}\left( {x + 5} \right) + 2 = \frac{x}{4} + \frac{{13}}{4}.$

See this picture

**tecknics** · September 16th 2009, 08:13 PM

Perfect! Thank you!

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