Question

Find the equation for the tangent line to the graph of the function f(s) = s3 at s = 1/2.
At what point(s) on the curve is the tangent line parallel to the line 2x − 3y = 7?

I got y-1/8=3/4(x-1/2), so none?

Answer 1

This stuff is going to be the death of me >>

Answer 2

2x - 3y = 7
-3y = -2x + 7
y = (2/3)x - (7/3)

At what points on y does the slope = 2/3?

3x^2 = 2/3
x^2 = 2/9
x = +- sqrt(2/9)
x = +- sqrt(2) / 3

At x = positive or negative sqrt(2) / 3, the tangent will have the same slope as the line given.

Answer 3

from there, i suppose we'd plug those x's back into y to find the y-value of the point

Answer 4

ahhh, this stuff is a pain >>

Answer 5

so if we plugged that x back into 2x-3y=7 it would give the y cord?

Answer 6

oh, i was not paying attention to the variables we were using, should be s.

plug those x values that should be s values back into the function f(s)

Answer 7

I was I knew how to navigate these functions I really need to study that. This is literally the only thing that is a foreign language for me

Answer 8

hmm... what exactly confuses you about these problems?

Answer 9

just what the equations mena

Answer 10

mean* like i just dont get the relationships it bugs the pellet out of me

Answer 11

Like, the relation between a function and its derivative?

Answer 12

how all 4 equations are related, I guess its just about working backwards I just dont get how that line, that were trying to find the parallel points, relates to anything....

Answer 13

If i were to get that on my test I'd get it wrong hands down.

Answer 14

The relation between the equation you find in the first question compared to the second question? For the most part, those questions are unrelated. The first question just wants you to find the tangent line where x=1/2. The second one wanted you to identify what points on f(s) had the same slope as the line they provided.

Answer 15

ahh and the derivative is the slope values so you set the derivative equal to the slope of the new line and got an x then if I wanted the y I would have to go back to f(s)

Answer 16

essentially its I'm finding a line the same way I would with a derivative

Answer 17

or the points rather

Answer 18

Yeah, pretty much.
You're finding the points on f(s) where the derivative (or in other words, the slope of the tangent at a point) is equal to the slope of that line.

Answer 19

I think I get it now thank you

Answer 20

At what point(s) on the curve is the tangent line parallel to the line 2x − 3y = 7?
2x - 3y = 7
-3y = -2x + 7
y = (2/3)x - 7/3

The slope of the line is 2/3. The slope of the tangent line of f(s) is 3s^2 for some value s. If the tangent line were parallel to this line, then the slope of the tangent would be equal to the slope of the line.
3s^2 = 2/3
s^2 = 2/9
s = +- sqrt(2/9)
s = +- sqrt(2)/3

f(sqrt(2)/3) = (sqrt(2)/3)^3
= 2sqrt(2)/27

f(-sqrt(2)/3) = (-sqrt(2)/3)^3
= -2sqrt(2)/27

At (sqrt(2)/3, 2sqrt(2)/27) and (-sqrt(2)/3, -2sqrt(2)/27 ), the tangent line is parallel to the given line.

\[ (\frac{\sqrt{2}}{3} , \frac{2\sqrt{2}}{27}), (-\frac{\sqrt{2}}{3},-\frac{2\sqrt{2}}{27} )\]