Expert: Scotto - 3/1/2009

Question
QUESTION: Hi,

I need help with this: the question is:

Determine the equation of the tangent y=1/2x^2-x-1/3 that is perpendicular  to y=8-3x. (The graph is added, plz check attachment)

This is the question it says to find it using derivatives, specially using the limits and power rule and sum and difference rule where applicable.


please reply soon, I really need Help soon

ANSWER: The slope of 8-3x is -3, so the slope of the perpendicular has the inverse of -3, which is -1/3.

To find where the graph has this slope, the derivative would be,
since f(x) is x²/2 - x - 1/3, f'(x) = x - 1.

This is perpendicular where x-1 = -1/3, or at x = 2/3.

Find f(2/3) to get y0 and take 2/3 as x0.

The equation of the line is then
y-y0 = m(x-x0) with m as the slope, which was -1/3.


---------- FOLLOW-UP ----------

QUESTION: Thank you I got that, I tried it this way. And I got 1/3 as my slope and if I put that into my equation, which I have to do and I get this. Is that correct: y=1/3x-11/9 plz reply..Thank you so much for your time to reply back. I really do appreciate it =P

Answer
I've seen people close the comment with a smiley face, but what I see in yours is a =P.  Maybe that was just a mistake, but we're not
really worried about that.

The function you gave has the right slope,
but does it hit the curve at that point?

Take x=2/3 and put it into f(x).
We see that f(2/3) = 4/18 - 2/3 - 1/3 = 2/9 - 1 = -7/9.

What this says is that the line is y - 7/9 = -(x - 2/3)/3.
Multiply it out and you get -x/3 + 4/9 +7/9 = -x/3 + 11/9.

Yes, you're right.  Does that give you the thrill of victory?