Question

Limits Calculus
1) find slope of the tangent line to the graph of f at the given point.
2) find the slope-intercept equation of the tangent line to the graph of f at the given point.

Problem #1: f(x)=-3x+4 at (1,1)
Problem #2: f(x)=3x^2 at (1,3)

Please help me, I found the answers but I'm not sure if they are right :\ THANKS!

Answer 1

for #1 \(f(x)=3x+4\) you have a line
the slope is a constant, namely \(3\)

Answer 2

doesn't matter what the point is for the first one, the slope is always 3

Answer 3

for the second one if
\[f(x)=3x^2\] then
\[f'(x)=6x\] if you want the slope at \(x=1\) find
\[f'(1)\]

Answer 4

I know but I don't think I got the right answer.

Answer 5

at a point of figuring, I got -3-3h+4-1

Answer 6

for the first one or the second one?

Answer 7

First

Answer 8

the first one is a line
you do not need a limit
the slope of a line is constant

Answer 9

if you are trying to compute
\[\frac{f(1+h)-f(1)}{h}\] we can do that too

Answer 10

oh and you get exactly what you wrote only divided by \(h\)

Answer 11

\[\frac{ -3-3h+4-1}{h}=\frac{-3h}{h}=-3\]

Answer 12

Oh I see!

Answer 13

but in any case it is the slope of \(y=-3x+4\) which is \(-3\)

Answer 14

For the second one, I did 3(1+h)^2-3(1)^2/h

and I got 3(1+2h+h^2)-9
then 3+6h+3h2-9
is that right?

Answer 15

3h^2 I mean

Answer 16

\(3\times 1^2=3\) not \(9\)

Answer 17

numerator should be
\[ 3+6h+3h2-3\]

Answer 18

which is
\[6h+3h^2\]

Answer 19

divide by \(h\) and you get
\[6+3h\] then let \(h\to 0\) and get \(6\)

Answer 20

Ah, I messed up on the 3(1)^2 and didn't notice my mistake.

Thank you so much for helping! Medal for you!

Answer 21

yw