# Tangent's equation

1. Nov 2, 2005

### Izekid

Decide an equation for the tangent to the curve y=6x-x^2 at the point (4,8)

Well for this I thought I'd use this

6*4-4^2=8

6*8-8^2=-16

Then I thought I'd Use

DeltaY/DeltaX

8-(-16)/8-4 =6

Then I thought the answer was -6X - 16 But it Wasen't the answer should be
-2x -16

What am I doing wrong please help me!?

2. Nov 2, 2005

### HallsofIvy

Yes, but you didn't need to do that- you were told that when x= 4, y= 8. That's what "at the point (4, 8)" means.
NO! (4, 8) are not two different x- values: they are the (x,y) values of a single point.
Basically, you are doing everything wrong! What you found was the equation of the line between the points (4, 8) and (8, -16).
You were asked to find the equation of the line through the single point (4, 8) that is tangent to the curve there.

I assume this is a calculus course, since otherwise there is no way to answer this question- but you didn't use any calculus!

Do you know how to find the derivative of y with respect to x and evaluate it at x= 4?
Do you know how "slope of the tangent line" is related to the derivative?

3. Nov 2, 2005

### Izekid

That was no Help

Yeah, I did know that the point was x1 = 4 And y1 = 8
And I know I decide those things with f(x+h)-F(x) / h and so on but I doesn't help me nada. And your quotation didn't help me either please provide me a solution so maybe I understand?

4. Nov 2, 2005

### TD

Have you seen derivatives yet?

5. Nov 2, 2005

### verty

Izekid, you need to tell us about derivatives.

6. Nov 2, 2005

### Izekid

You shall not know the derivate for the equation.... !
Its a simple question where you shall DECIDE(you shall tell the one who does the question about what the equation shall be) the equation for the curve 6x-x^2
And you have 2 coordinates (4,8)

And The Answer Should be Y=-2x-16

But if you want to derivate this it's gonna be
X^2 = 2x^2-1
=2x

6x - 2x = 4x And that's no answer

So please help I have written the things that is written in my math book...

7. Nov 2, 2005

### finchie_88

Let f(x) = 6x-x^2
therefore f'(x) = 6-2x. At point (4,8), f'(x) = -2. So, we say that the following is true...
8-y = -2(4-x) => y = 16-2x (as required, I think)

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