Question

Find equations of the tangent line and the normal line to the graph of f at P.

y = 2x - (4 \ sqrt(x))

Answer 1

\[y = 2x - (4\div \sqrt{x})\] and P(4,6)

Answer 2

\[y \prime= 2-4(-1/2x ^{-3/2})\]

Answer 3

I'm sorry, how are you getting the differential to be that answer?

Answer 4

the derivative of \[1\div \sqrt{x}=x ^{-1/2}=-1/2x ^{-1/2-1}=-1/2x ^{-1/2-2/2}=-1/2x ^{-3/2}\]

Answer 5

we can factor out the 4 in the numerator and multiply it tho our previous result to get
\[-2x ^{-3/2}\]

Answer 6

to find the slope of our tangent line we can substitute (4,6) into the equation to get
y'=2+2(4)^(-3/2)

Answer 7

let me know if you don't understand anything so far.

Answer 8

Are you sure this is correct? The tangent line slope is 9/4 according to my book.

Answer 9

2+2(4)^(-3/2)=2+2(1/8)=2+1/4=8/4+1/4=9/4
i wanted to give you the general equation so you could substitute 4 for x.

Answer 10

now that you have the slope of your tangent line, solve for b using slope intercept form and substituting x=4 and y=6 like so:
y=mx+b where m=slope
6=9/4(4)+b
6=9+b
b=-3
therefore, the equation of your tangent line is y=9/4x-3

Answer 11

Now I see it! Thanks a bunch! By using the "slope of the tangent line" equation, you got the slope 9/4. You simply substitute in the Point (4,6) for the x- and y-values and get the "b" value. Then you display the equation!
Brilliant...

Answer 12

The slope of the tangent line being m = lim (as h approaches 0) f(a+h) - f(a) divided by h.

Answer 13

Exactly! Let me know if you need any more help :)