Question

find the slope of the tangent line

Answer 1

to the parabola y=4x - x^2 at the point (1, 3) using
\[m=\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]
and
\[m=\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\]
then find an equation of the tangent line.

Answer 2

for @shamil98 only!

Answer 3

can i?

Answer 4

No idea .

Answer 5

lol jk

Answer 6

...

Answer 7

The tangent to a curve at a point is the 1st derivative at that point

Answer 8

and your point is?

Answer 9

y=4x - x^2
dy/dx = 4 - 2x
y = (dy/dx) x + b
place the values of the point into equation and solve

Answer 10

I don't like those ugly dy/dx notations

Answer 11

Neither do I, I'm an old school y' er.

Answer 12

The were developed in the foundation of Calculus

Answer 13

not elegant for my taste

Answer 14

I understand your development but why reinvent the wheel

Answer 15

the foundation of calculus did not use dy/dx
i doubt the early greeks nor the early egyptians used it

Answer 16

I guess my point is that I've given the definition to which the slope of the tangent line will be solved, and yet you two insisted on using dy/dx

Answer 17

Your both definitions are one and same.
you can jump from first definition to second definition by substituting x-a = h

so i think finding the slope by using one definition will do

Answer 18

you're missing the point >.< too

Answer 19

I bet you three are the Einstein of your country... but we're dealing with future Richard Feynman here

Answer 20

yes whats the point :o

Answer 21

im just feeling dumb right now lol

Answer 22

laughing out loud

Answer 23

\[y=-x^2+4x @(1,3); m=\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]
\[m=\lim_{x \rightarrow a}\frac{ -x^2+4x-[-(1)^2+4(1)] }{ x-1 }\]
\[m=\lim_{x \rightarrow a}\frac{ x^2+4x-3 }{ x-1 }\rightarrow m=\lim_{x \rightarrow a}\frac{ (-x + 3)(x-1)}{ x-1 }\]
\[m=\lim_{x \rightarrow a}-x+3=-(1)+3=2\]

Answer 24

some textbooks refer to that definition as the "alternate-form"

Answer 25

we're taught h->0 definition i guess. but the second one is not that hard to *see* once u digest the first def

Answer 26

that is the second part @rsadhvika

Answer 27

^ I see your point now :)

Answer 28

because later on that h will turn into delta x

Answer 29

ohk i get it completely. basically you want to go through it step by step + slow --- so you see everything; Nice :)

Answer 30

\[y=-x^2+4x@(1,3);m=\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\]
\[m=\lim_{h \rightarrow 0}\frac{ [-(a+h)^2+4(a+h)]-[-(1)^2+4(1)]) }{ h }\]\[m=\lim_{h \rightarrow 0}\frac{ -(1)^2-2(1)h-h^2+4(1)+4h-3 }{ h }\rightarrow \frac{ h^2+2h }{ h }\]
\[m=\lim_{h \rightarrow 0}\frac{ -h(h-2) }{ h }\rightarrow m=\lim_{h \rightarrow 0}-1(0-2)=2\]

Answer 31

the tangent line to y=f(x) at (a, f(a)) is the line through (a, f(a)) whose slope is equal to f'(a), the derivative of f at a
\[f'(a)=2; y-f(a)=f'(a)(x-a)\]
\[y-3=f'(a)(x-1)\rightarrow y-3=2x-2\rightarrow y=2x+1\]