Finding the equation of the tangent line. Please help!

[thomcart8]

Ok this is the equation I had to find now all i have to do is sketch the graph, but i got a little stuck.
P=30e^(-3.23x(10^-5)h)

Ok h is feet above sea level. The P is the air pressure in inches of mercury. And air pressure at sea level is 30 inches of mercury.
Ok what is the equation for the tangent line at h = 0. I know what the graph looks like it is a line going down though x=0 y = 30.

 

[galactus]

Find the derivative of \(\displaystyle P(h)=30e^{-.0000323h}\)

Then, sub in h=0 to find the slope, m, at h=0

Sub h=0 into P(h) and find the corresponding y value.

Sub them all into y=mx+b and solve for b.

You're finished.

If you graph this, the slope is so slight that it appears as a horizontal line crossing at y=30.

But, it does have a small slope.

 

[thomcart8]

thanks