Question

(a) find an equation of the tangent line to the
graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.

Answer 1

\[f(x)=x^2+2x+1 \]

Answer 2

(-3, 4)

Answer 3

you wanto find the tangent line thru (-3, 4)

Answer 4

you need to knw slope to write the equation of tangent,

slope of tangent at x = -3, is \(f'(-3)\)

Answer 5

start by finding \(f'(3)\)

Answer 6

\(\large f'(3) = \lim \limits_{x \to 3} \frac{f(x) - f(3)}{x-3}\)

Answer 7

\[f(3)=3^2+2*3+\]

Answer 8

yes do it all the way in ur notes,
and tell me wat u get for \(f'(3)\)

Answer 9

\[f(3)=3^2+2*3+1\]

Answer 10

\[f(3)=9+6+1, f(3)=16\]

Answer 11

yes, plug it above

Answer 12

\[f'(3)=\lim_{x \rightarrow 3}\frac{ f(x)-16 }{ x-3 } \]

Answer 13

yes, keep going

Answer 14

i'm stuck :(

Answer 15

wait no

Answer 16

:)

Answer 17

\[f'(3) = \lim_{x \rightarrow 3}\frac{ (x^3+2x+1)-(16) }{ x-3}\]

Answer 18

Yes !

Answer 19

wat next

Answer 20

\[f'(x)=\lim_{x \rightarrow 3}\frac{ (x^3+2x-15) }{(x-3) }\]

Answer 21

factor numerator and see if it cancels the bottom illegal thingy :)

Answer 22

3-3=0.

Answer 23

hey, how come x^2 became x^3 :o

Answer 24

woops sorry, it should be x^2

Answer 25

\(\large f'(x)=\lim \limits_{x \rightarrow 3}\frac{ (x^{\color{red}{2}}+2x-15) }{(x-3) } \)

Answer 26

Yes :) try to factor numerator

Answer 27

(x(x+2))-15

Answer 28

no not like that

Answer 29

\(x^2+2x-15\)
think of two numbers such that

sum = 2
product = -15

Answer 30

5-3 = 2
5*-3 = -15

so,
\(x^2+2x-15 = (x+5)(x-3)\)

Answer 31

oh yeah

Answer 32

plug them above, and tell me wat u get for final f'(3)

Answer 33

f'(x)=limx-->3 x+5

Answer 34

Yes, now that the bottom illegal thing is gone, you can take the limit now
plug x = 3

Answer 35

f'(x)=limx-->3 x+5
= 3+5
= 8

Answer 36

f'(3)=8

Answer 37

Yes, f'(3) is the slope of given function at x = 3

so you knw the point and u knw the slope

Answer 38

you can write the equation of line in point-slope form

Answer 39

Should I be putting f'(x) in front of the lim or f'(3)

Answer 40

Very good question :)

Answer 41

since we're finding the derivative AT x = 3
we should put f'(3) oly

Answer 42

if you're finding derivative in general for ANY X, then we put f'(x)

again, its just a notational thing

Answer 43

so, wats the equation of tangent line ?

Answer 44

y-4=8(x--3)
y-4=8(x+3)

Answer 45

Awesome :)

Answer 46

So that is the answer for A?

Answer 47

yes

Answer 48

Ok for b can i just put that equation in GeoGebra to get the right graph

Answer 49

Yes ! put the equation for f(x) in geogebra

Answer 50

plot the point (-3, 4) also

Answer 51

What should I label the point?

Answer 52

A = (-3, 4)

Answer 53

or

P = (-3, 4)

Answer 54

\[f(x)=x^2+2x+1 \]

Answer 55

anything wil do

Answer 56

Put that in geogebra?

Answer 57

yes !
plugin both funciton and point

Answer 58

and graph the tangent line which u got also : y-4=8(x+3)

Answer 59

graph below 3 :-
1) function f(x) = x^2+2x+1
2) point P = (-3, 4)
3) tangent y-4=8(x+3)

Answer 60

how do i use the derivative feature to check

Answer 61

looks good, let me open geogebra and try if we can plot the derivative function

Answer 62

will my teacher know if i checked it or not?

Answer 63

she wont, but if u have time... it is goot to confirm girly :)

Answer 64

lol don't worry about it i trust you

Answer 65

:) once i get to knw how to do it in geogebra, il let u knw...

Answer 66

thanks!