Find the slope of the curve.
the slope at a specific piont or the whole curve?
y=x^{2}-4x at x = 1
ik you use lim h->0 \[\frac{ f(a+h)-f(a) }{ h } \]
you could also take the derivative dy/dx and evaluate at x = 1
but your definition and approach should work also... I just don't recall that one as well. But it's much more general, I think... can be good to know.
i havnt done derivitive yet i have to do it that way ^^^ but i need help
oh, sorry :(
thx anyways
your equation is finding the slope at a specific point of the curve, the derivative if for the entire function
like i said above... im not doing a derivitive
oops, I messed that up...
ooo never mind ik how to do it now. thx!
f(a) is just like f(x). f(a + h) is like f(x+a)... just substitute in (a+h) for every x term. Then subtract the expressions and divide by h, and take the limit as h ->0
\[\frac{ (a+h)^2 +4(a+h) - f(-1) }{ h }\]
for what it's worth, the "meaning" of your limit expression means, compare the value of the function at two points, one with x value of "a" and the other separated by h units, are "a+h". Now shrink that h value smaller and smaller toward h-->0. The amount of change in the value of the function when h is approaching 0 is the slope at that point x=a
I think that middle term should have been -4(a+h) Also, leave f(a) in the last term as (a^2 - 4a) (don't sub in x=1 yet)
the slope equals 3
\[\frac{ a^2+h^2+ah+4a+4h+3 }{ h }\]
plug in -1 for a everywhere
\[\frac{ 1+h^2-h-4+4h+3 }{ h }\]
\[\frac{ h^2-h+4h }{ h }\]
limh h->0 h+3=3
nice :)
thanks haha
i have to do this pellet A LOT
bahahah i didnt type pellet
im about to post another. you have to do it the same way but im not sure what to do next
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