Question

Find eqn of tangent line to graph of f(x)=4/√x @ (1,4)??
I am stuck on simplifying after plugging in to formula

Answer 1

when you know a point and slope; how do you form a line?

Answer 2

(Px,Py), m

y-Py = m(x-Px)

m = slope = first derivative of f(x)

Answer 3

so slope is -2x^(-3/2)?
i was plugging in to the 'limit as delta x goes to 0' formula

Answer 4

why would you ever wanna use the limit version? ewww

Answer 5

f(x)=4x^(-1/2)

f'(x) = -2x^(-3/2) is correct; now when x=1 we got our slope

Answer 6

ok, so f' = -2x/x^(3/2)?, then?

Answer 7

you got an extra x in that ... prolly a typo

Answer 8

since we want to know the slope at the point (1,4), plug in for x=1

Answer 9

i mean -2 for numerator

Answer 10

1^(...) = 1; so we are left with m=-2

Answer 11

(Px,Py), m
(1 , 4) , -2

y-Py = m(x-Px)
y- 4 = -2(x-1)

and rearrange to your hearts content

Answer 12

into y=mx+b ?

Answer 13

ax+by=c

y-Py=m(x-Px)

ax/c + by/c = 1

y = mx -mPx + Py

y = mx + b

any form you want to use ....

Answer 14

a line is a line is a line regardless of how you want to shape its equation

Answer 15

i'm sorry the question asked for slope int. form thanks!

Answer 16

then yes, reshape it into the y=mx+b ;)

Answer 17

y- 4 = -2(x-1)

y- 4 = -2x +2
+4 +4
-------------
y +0 = -2x+6


y = -2x+6