Well, seeing as I will be working on calculus for a long time, and I figure at the rate I've been going I'll get stuck a lot and need help, I'll just make one thread for all my calculus troubles.(adsbygoogle = window.adsbygoogle || []).push({});

I just read the chapter on Tangent Lines and Slopes. I did a problem, double-checked the math, then turned to the back of the book and found out the answer was completely incorrect. I retraced again, and found no problem, but the book is a bit confusing so I might have missed something.

The problem reads: Find a formula that gives the slope at any point P (x,y) on the given curve. [itex]y=4-x^2[/itex] [itex]P:(1,-1)[/itex]

My work is as follows:

1)I begin by making a secant line from point P to a point Q, which I arbitrarily place. The coordinates are [itex]Q:(-1+\Delta x, 4-(-1+\Delta x)^2[/itex]

2)To find my slope: [tex]\frac{\Delta y}{\Delta x} = \frac{4-(-1+\Delta x)^2}{\Delta x}[/tex]

3)This becomes: [tex]\frac{4-1+2\Delta x-\Delta x^2}{\Delta x}[/tex]

4)Finally, I get [itex]5-\Delta x=m[/itex] Getting rid of the [itex]\Delta x[/itex] my slope is [itex]m=5[/itex]

The back of the book says the answer is -2x.

What happened?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# All my Calculus Troubles

**Physics Forums | Science Articles, Homework Help, Discussion**