Find an equation of the tangent line to the curve
- Thread starter Lividtea
- Start date
The function is
y(x) = x3 - x2 + 3
and you have computed the derivative and slope correctly. However the (x1, y1) in the tangent line formula is for the point (2, y(2)) so where did you get the zero?
In this question, I mistakenly thought x=2 is the same as (2,0). How were you able to interpret the points to be (2, y(2))? And yes I was able to get the correct answer, thank you~
Bob Brown MSEE
Full Member
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- Oct 25, 2012
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- 597
No derivative required
Translate so that the orign is at the point currently on (2,7)
(y+7) = (x+2)^3-(x+2)^2+3
y = x^3+5x^2+8x
Truncating all but linear
y = 8 x
Translate origin back where it was originally
(y-7) = 8 (x-2)
Answer is,
y = 8x-9
Translate so that the orign is at the point currently on (2,7)
(y+7) = (x+2)^3-(x+2)^2+3
y = x^3+5x^2+8x
Truncating all but linear
y = 8 x
Translate origin back where it was originally
(y-7) = 8 (x-2)
Answer is,
y = 8x-9
Last edited:
Steven G
Elite Member
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- Dec 30, 2014
- Messages
- 14,603
every point on the line is (x, y(x)) 0r if you prefer (x,f(x)). You are giving two pieces of information: y=x^3 -x^2 + 3 and x=2. Since a point has an x-value AND a y-value and you have the x-value (x=2) you need to find the y-value. So use the equation y=x^3 -x^2 + 3 and plug in 2 for x.
You thought that if x=2 then y=0. Although this is wrong let's go with it for now. In your work you replaced x with 2 and y' with 0. y and y' are not the same! So if y=0 there is no reason to assume y' =0 as well. Be careful with this!