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cute_but_crazy

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Feb 11, 2006
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ok so the question was...... Find teh equation of teh tangent line to the graph of the given function f(x)=4-x-3x^2 given teh point (-1,8).

So I entered it into teh (f(x)-f(a))/x-a formula and I got 5

then I entered it into the (f(a+h)-f(a))/h formula and got -3

so yah.....

and I forgot to mention teh answer is 7
 
Re: Identitie

What exactly was the question?

I might guess it was to find the equation of a tangent to f(x) passing through (-1, 8), but your work is somewhat dodgy: points on f(x)? numerical outputs?

Be more specific as to what you're doing.
 
sorry I was giving teh answers for teh slope but here's teh question.......

Find the equation of the tangent line to the graph of teh given function at teh given point.
f(x)=4-x+3x^2, (-1,8)



and there is two equations to find the slope. I used both and got different answers
 
Thanks for clearing that up. However, I still don't know how you got those values.

The second version of your formula is probably easier to work with.

\(\displaystyle \mbox{ f(a) = 3a^2 - a + 4}\)

\(\displaystyle \mbox{ f(a + h) = 3(a + h)^2 - (a + h) + 4 }\)
\(\displaystyle \mbox{ = 3a^2 + 6ah + 3h^2 - a - h + 4}\)

So\(\displaystyle \mbox{ f(a + h) - f(a) = (3a^2 + 6ah + 3h^2 - a - h + 4) - (3a^2 - a + 4)}\)

Simplifying this and substituting into your formula (as a limit as h tends to 0) may be an improvement.

The slope at (-1, 8), that is, f'(-1), should be -7, not +7.
 
thnk you so much .......I found out where I went wrong....I never distributed teh 3 throughout teh quadratic formula....lol
 
Do the same as we did in the other one.

g(x + h) and g(x) are two fractions which can be subtracted by cross multiplying, and you will get some cancellation by expanding from there.

Show all your work.
 
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