Question for skeeter or whoever...

[tkthustler]

Still having trouble... A couple of days ago I asked this question... Find the equation of the line tangent to the graph of f(x) = 2x^2 - 6x + 1 at x =1 I recieved this answer....

*********Find your derivative.

That gives you the slope at a point, in this case, x=1. 4(1)-6. The slope at

x = 1 is -2.

Your y value at x = 1 is y = 3. So, you need the slope of the tangent line at (1, 3)

You have the slope(m) and x, y coordinates(1 and 3). Use y = mx + b to find b, then you have your line equation*******

I still dont understand though.. I know by finding the derivative and the answer from that is your slope which is -2. Buy isn't the Y value - 3 not 3? Then dont you plug that info into Y-y1= m(x-x1)?? Guess I'm still a little confused.

Also one more question. What is the quick way to solve this problem using the chain rule I believe. I checked my calk book but all they have is a bunch of letters and symbols. If someone could explaina couple of my example problems I would be most appreciative because I have a test on Monday. Thx

EXAMPLES

X(4-x)^3

x^4/(3X-8)^2

4X^3(3X-1)^8

 

[daon]

0) Find y when x is 1. y = 2(1)^2-6(1)+1 = -3. Your point is p=(1,-3).
1) Take derivative: y'=(2x^2 - 6x + 1)' = 4x-6
This is your slope at an arbitrary input x.
2) Take a specific x to find the slope at. You have x=1, thus y' = -2.
This means the line tangent to y=(2x^2-6x+1) when x = 1 has a slope of -2.
3) Use the formula you posted with your point p and solve for y.

I'm not sure what you mean by the "quick way" using chain rule, please be more specific if you could. The chain rule only applies when you have composite functions.

 

[galactus]

Hello tkhustler. I am the one who posted originally. I think you're just making it more difficult than it needs to be. Which is not uncommon.

Follow me and Daon's suggestions.

Find the derivative of your function: 4x-6. You have it that far. That is your slope.

You are asked to find the slope at x=1. Sub 1 into your derivative. That equals -2. That is the slope at x=1. See, so far?.

Sub x=1 into your function to find the corresponding y value.

\(\displaystyle \L\\2(1)^{2}-6(1)+1=-3\)

Now, you have x, y, and m. All 3 needed to find your tangent line.

\(\displaystyle \L\\y=mx+b\)

\(\displaystyle \L\\{-}3=(-2)(1)+b\)

b=-1

There's your equation: \(\displaystyle \L\\y=-2x-1\)

I, too, don't know what you mean by the quick chain rule method.

There's no chain rule here.

But the few you have at the bottom of your post would use the chain rule.

\(\displaystyle \L\\x(4-x)^{3}\)

Product rule:

\(\displaystyle \L\\(x)3(4-x)^{2}\underbrace{(-1)}_{\text{chain rule,\\derivative of 4-x}}+(4-x)^{3}(1)\)

=\(\displaystyle \L\\{-}3x(4-x)^{2}+(4-x)^{3}\)

Factor out \(\displaystyle \L\\(4-x)^{2}\):

\(\displaystyle \L\\(4-x)^{2}(-3x+(4-x))\)

\(\displaystyle \L\\(4-x)^{2}(-4x+4)\)

Factor out -4:

\(\displaystyle \L\\{-}4(4-x)^{2}(x-1)\)

 

[tkthustler]

Thx so much