Question

Find an equation of the tangent line to the curve at the given point.
y = sqroot(x)
Point: (64, 8)

Answer 1

\[y-8=f'(64)(x-64)\]

Answer 2

i need to use equation f(x) - f(a) /x-a to find slope first but im confused on what f(a) would be

Answer 3

because f(x) would just be sqroot(x) right?

Answer 4

right

Answer 5

do you know deriviative rules yet?

Answer 6

also i meant the lim as x approaches a for that equation...and slightly, derivatives are what we are just starting

Answer 7

1/2SQROOT(x)
1/2(8)
1/16

Answer 8

\[\frac{\sqrt{x}-\sqrt{a}}{x-a}\]
\[\frac{\sqrt{x}-\sqrt{a}}{x-a}*\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\]
\[\frac{(x-a)}{(x-a)(\sqrt{x}+\sqrt{a})}\]
\[\frac{1}{\sqrt{x}+\sqrt{a}}\]

Answer 9

when x = a we have the slope at x

Answer 10

im a little confused ha...so a would equal x?

Answer 11

\[\lim_{a->x}\frac{1}{\sqrt{x}+\sqrt{a}}=\frac{1}{2\sqrt{x}}\]

Answer 12

y = (x)^1/2 ,y '=1/2sqrtx, using the point (64, 8)
at x=64, y '=1/2sqrt64=1/16
y-64=(1/16)(x-8)
y= (x/16)+4

Answer 13

yes, when a = x is what we get when as a gets to x

Answer 14

so it would be sqroot(a)

Answer 15

not if you follow the algebra :)

Answer 16

1/2sqrt(a); but it makes more sense in the problem to say that a approaches 64 until a = 64

Answer 17

and since x=64 ....

Answer 18

im lost haha...i dont understand

Answer 19

would a = 64? 2(sqroot(8)) ?

Answer 20

er 2(sqroot(64))

Answer 21

thanks i worked out the problem