Find the equation of the tangent to $y=x^3-2x^2+1$ in the form $y = mx+c$.

Question

For $P = (1, 0)$

First I would differentiate,

$$y=x^3-2x^2+1$$

To get

$$\frac{dy}{dx}= 3x^2-4x$$

Then add in 1 for x.

= $-1$

But then I am really not sure where i should do with this to get the answer

$$Y=1-x$$

How to I reach this point?

Yes sorry that was from a wrong example. Question still stands though. — jackdh, May 06 '13 at 17:20

score 1 · Accepted Answer · answered May 06 '13 at 17:35

1

You have the point $(x_1, y_1) =(1,0)$ and the slope $m = -1$.

You can use the so-called "point slope formula": $$y-y_1 = m(x-x_1)$$ $$y - 0 = -1(x-1)$$

And simplify.

answered May 06 '13 at 17:35

The Chaz 2.0

Thank you that rung the bell i was looking for !! – jackdh May 06 '13 at 17:36
1

Might I suggest that you go back to basics: ask yourself "what am I supposed to find?" (Ok, I need to find the equation of a line...) "What information do I need to write an equation of a line?" (Ah yes... a point and the slope) – The Chaz 2.0 May 06 '13 at 17:37

score 0 · Answer 2 · answered May 06 '13 at 17:18

0

Hint: 16 is not the value of $\frac{dy}{dx}$ at $x=1$.

answered May 06 '13 at 17:18

vadim123

2 Answers2