Equation of Tangent.

[AustrianSaurkraut]

For this question, I tried to begin it but kept getting stuck with the numerator term. I'm unsure what to do from here, because the bottom denominator is not factorable. Would I use quotient rule by turning the √x to x^1/2? But then I am left with a negative exponent. Any help will be greatly appreciated.


Determine the slope of the tangent to the curve y= 55-√x/x²-3x-12 at the point where x = 4. EXACT VALUES ONLY, AS SIMPLIFIED AS POSSIBLE

 

[skeeter]

[MATH]y=\frac{55-\sqrt{x}}{x^2-3x-12}[/MATH]
[MATH]y' = \frac{(x^2-3x-12) \cdot \left(-\frac{1}{2\sqrt{x}}\right) - (55-\sqrt{x})(2x-3)}{(x^2-3x-12)^2}[/MATH]
now determine [MATH]y'(4)[/MATH] and finish ...

 

[AustrianSaurkraut]

[MATH]y=\frac{55-\sqrt{x}}{x^2-3x-12}[/MATH]
[MATH]y' = \frac{(x^2-3x-12) \cdot \left(-\frac{1}{2\sqrt{x}}\right) - (55-\sqrt{x})(2x-3)}{(x^2-3x-12)^2}[/MATH]
now determine [MATH]y'(4)[/MATH] and finish ...

Thank you, I miswrote the question it was actually 55+ √x, making the equation 55+√x/x²-3x-12 Would this sign change affect the solution you showed by a lot. How would the new one look? Also thank you for the help.

 

[skeeter]

How would the new one look?

change the two signs affected ... surely you can do that, no?

 

[lookagain]

AustrianSaurkraut, make sure you use grouping symbols when writing this "expression" in horizontal style:

. . . it was actually 55 + √x, making the function/expression (55 + √x)/(x² - 3x - 12 )