Question

Find the equation of the tangent line to the curve.

Answer 1

\[y=x^3+2x^2+5x-3 at x=5\]

Answer 2

at x=5

Answer 3

So a tangent line touches a curve at one point, right? Can you tell me what point will the tangent line touch? (x,y)

Answer 4

find the derivative of y, and solve it using x =5

Answer 5

Y U OFFLINE?????

Answer 6

y=3x+4x+5 x=5

Answer 7

@coolsday

Answer 8

Now find the slope when x=5

Answer 9

i dont know how to

Answer 10

the derivative function is also called the slope function.

Answer 11

the derivative function can be used to find the slope of the original function at any given x - value. So what is the slope at x=5?

Answer 12

can you show me step by step....im not understanding

Answer 13

f'(x)=3x+4x+5

f'(x) is the slope at the given point

and you are required to find the slope at x=5

Answer 14

so you replace the x for 5

Answer 15

yes, and you differeintiated ur function wrong. Check again.

Answer 16

f'(x)=3x+4x+5
f'(x)=3(5)+4(5)+5
f'(x)=40

Answer 17

Yes, that is the correct way to do it

However, your derivative function is wrong

Answer 18

what am i doing wrong....i dont see any error

Answer 19

what is the derivative of x^3?

Answer 20

3x^2

Answer 21

why did u write 3x?

Answer 22

oh........haha small mistakes.. i must be careful

Answer 23

yeah, once you find your slope, you have to find the point the slope touches the original function. Can you find the point? (x=5, y=?)

Answer 24

x=5 y=40

Answer 25

y is not 40. You know that the tangent line touches a point on the curve when x=5. The equation of the curve is also given to you in the question.

Answer 26

so whats the 40 for?

Answer 27

now im getting confuse can you just show me

Answer 28

step by step

Answer 29

Let us find the point at which the tangent line touches at okay?

Given the equation f(x)=x^3+2x^2+5x-3, substitute x=5 into this equation to find the y value of the point

What is the point that the tangent line touches at?

Answer 30

37

Answer 31

nopeeee.. try again

Answer 32

wait

Answer 33

197

Answer 34

yes, so the point the tangent line touches the curve is at (5,197)

Answer 35

oh okay now i understand that

Answer 36

Now lets find the slope of the tangent line at that exact point (5,197)

Take your derivative function and find the slope when x=5.

What is the slope?

Answer 37

HOLY CRAP! X WAS SUPPOSE TO BE 2 NOT 5

Answer 38

.....

Answer 39

well, doesnt matter. repeat the previous step for x = 2

Answer 40

OKAY HOLD ON

Answer 41

2,27

Answer 42

check ur calculations again... it should be (2,23)

Answer 43

nope i dont get 23

Answer 44

do you replace the 2 on the original equation or the derivative equation

Answer 45

replace the x i meant

Answer 46

orginal function.

Answer 47

ohh okay

Answer 48

so do u get (2,23)?

Answer 49

now i got 23

Answer 50

ok, so (2,23) is the point on the curve where the tangent line touches at

Answer 51

Now lets find the slope of the tangent line at that exact point (5,197)

Take your derivative function and find the slope when x=5.

What is the slope?

Answer 52

i mean when x=2

Answer 53

not x=5

Answer 54

and the point should be (2,23) and not (5,197).

Answer 55

3x^2+4x+5

Answer 56

yes, what is the slope when x=2?

Answer 57

what is f'(x)?

Answer 58

i just said it...3x^2+4x+5. do i need to substitute x for 2 in the derivative

Answer 59

yes because u want to find the slope, f'(x)

Answer 60

i meant what is the VALUE of f'(x)?

Answer 61

when x=2

Answer 62

25

Answer 63

ok, so the slope of the tangent line at x=2 is 25.

let m=slope. So m=25. We also know the point (2,23).

Its basic algebra now. Can you find the equation of the tangent line?

Answer 64

Hint* line equation hint* f(x)=mx+b Hint*

Answer 65

y=25x-27

Answer 66

congratz! You got it!

Answer 67

yayyyyy!!!!!!!!!!!!!!!!! omg thank you very much

Answer 68

you stayed with me till the end, not many people do that

Answer 69

yeah, no problem :)

Answer 70

Well... I got to sleep now. So goodnight!

Answer 71

goodnight