Two Tangent Line Problems?

Question

I'm having trouble with these. One guy tried to explain it to me but I still couldn't get the right answer... Help please!

Find the equation for the tangent line to the graph of f(x) = 2x^2 - 2x + 3 at the point where x=1

Find all points on the graph of f(x) = -x^3 + 3x^2 - 2 at which there is a horizontal tangent line.

Answer 1

First thing to do is to find the gradient of the tangent - by differentiation.

f'(x) = 4x - 2 so when x = 1 f'(x) = 2

Then we find the location on the curve i.e f(1) = 2-2+3 = 3
So we know that the tangent has a gradient of 2 and passes through the point 1,3

Consider any point (X,Y) on the tangent. The gradient (which we know) is defined as the change in Y / change in X

i.e. (Y - 3) / (X- 1) = 2
rearrange Y- 3 = 2X - 2
or Y = 2X + 1
or Y - 2X - 1 = 0 depending on which version you want.

For the second part if the tangent is horizontal then the gradient is zero. so f'(x) = 0

Differentiating

-3x^2 + 6x = 0

-3x (x-2) = 0
so x = 0 or x = 2
x = 0 y = -2
x = 2 y = 2

Answer 2

Take the derivative of the function f(x) as this will give the slope of the tangent
So f'(x) = 4x - 2
Plug in the value x=1
So slope = 2.
Next plug in the value x = 1 back in the original equation
that is f(x) = 2-2+3 to get the y coordinate.
Now this x, y point is common to the graph as well as the tangent. So using the formula y-y1 = m(x-x1) where m is the slope of the tangent(which you computed previously). and y1 and x1 are the points common with the graph and the tangent, you can find the equation of the tangent.

Part 2
basically take the deriavative again of f(x) and solve that equation (ie. equate it to zero cause it is a horizontal tangent line) and solve it.to get x.

Answer 3

I'll do the first (only one answer per question).

Find the derivative: f'(x)=4x-2. Plug x=1 into this to find the slope of the tangent line: m=4*1-2=2. Now find the y-value for this function at x=1: y=f(1)=2*1^2-2*1+3=3. So you want a line with slope 2 that goes through the point (1,3).

THis gives y-3=2(x-1), so y=2x+1.

Answer 4

At the point x=1, the slope of f(x) is given by f`(x)=4x-2, so f`(1)=2. Now just use the point slope method: y-f(1)=2(x-1), but since f(1)=3, the formula becomes y-3=2(x-1).

Answer 5

For 2a. f"'(4) is -12 b/c x/6= -2 2b. P2(x) = -3 +10(x-4) -6(x-4)^2 (you will desire to differentiate it) 2c. combine P(x), yet use t, then replace t with x and you get: 7x-3/2(x-4)^2 + 5/3(x-4)^3 -a million/2(x-4)^4 - 28 (the 28 is once you plug in 4) i'm undecided approximately d.