Question

Find the equation of the tangent line to the curve f(x)=x^2+5x at x=2

Answer 1

You are familiar with derivatives I hope? :-)

Answer 2

:)

Answer 3

I am trying but I am really bad at math!

Answer 4

I'm not even sure where to start with this one..

Answer 5

First, you'll want to take the derivative of f(x)

Second, plug in x=2 into the new function f'(x) you just found.

Third, take that slope (call it 'm') and put into the point-slope formula which looks like this:
\[y-y_0=m(x-x_0)\]

Fourth, put x=2 into your original function f(x), to find x_0 and y_0, your coordinates of your tangential point where the tangent line just touches your parabola

Finally, put in everything into your point-slop formula and make it look like y = mx + b, tada!


If you get stuck give a shout out for help :D But see if you can do it first, I think you can

Answer 6

The derivative gives you a function of the slope of your original function in terms of the input variable (x) in this case.

Answer 7

would the derivative be 2x+5

Answer 8

did you learn the shortcuts of taking derivatives yet? or are you supposed to find the derivative using the definition still?

Answer 9

I don't think that I have learned the shortcuts, do you mean like the power rule and things.

Answer 10

\[f'(x) = \frac{d}{dx} x^2 + 5x = 2x^{2-1}+5x^{1-1}=2x^1+5x^0=2x+5\]

Answer 11

Does that look familiar?

Answer 12

yes

Answer 13

What you'll do is input 2 for x into that function

Answer 14

Read the steps I posted above :-P

Answer 15

y=9?

Answer 16

yes, but more specifically that's 'm', your slope of the tangent line

Answer 17

That's the slope at that exact moment on your parabola at the point where x=2

Answer 18

hint for step #4:
\[x_0 = 2\]

Answer 19

y'=9 technically, sorry minor detail

Answer 20

woah, woah, woah, im now completely confused

Answer 21

I dont know where we are putting everything in this equation:
y−y0=m(x−x0)

Answer 22

You're doing fine! You found the slope, now put that sucker into your point slope formula and go find your x_0 and y_0, your coordinates

Answer 23

@jeanniebean , In your original function, what do you get as your output when you put in 2 for x?

Answer 24

ok so y=mx+b is y=2(2)+5 ?

Answer 25

Hold on, just answer the question I just asked for a moment. You'll be answering your own question

Answer 26

14

Answer 27

y=x^2+5x @ x=2 ... y=???

Answer 28

Yes!

Answer 29

So isn't the point (2,14) our coordinate?

Answer 30

oh!!!

Answer 31

\[(x_0,y_0)\]

Answer 32

plug & chug now into the point-slope formula. Go go go banana! :D

Answer 33

y−y0=m(x−x0) = y-14=m(x-2)

Answer 34

You have 'm' already remember? :P

Answer 35

is that 9?

Answer 36

Yep!

Answer 37

so y-14=9(x-2)

Answer 38

Distribute the 9, add 14 to both sides to move it over to the right, and you have....? (answer)

Answer 39

how do you distribute it to both sides?

Answer 40

oh never mind... sorry read it wrong

Answer 41

I was going to say! lol

Answer 42

:-P sheesh (jk jk)

Answer 43

So do you have you answer now?

Answer 44

y-19x - 4 ?!?

Answer 45

*sigh* no...

\[y-14=9(x-2)\]
\[y-14=9x-18)\]
\[y=?\]

Answer 46

9 not 19... y=9x - 4 ?!?

Answer 47

Tada!!! You did it! gj

Answer 48

yay thanks soo much!

Answer 49

those two intersect at EXACTLY (2,14) where the slope is 9 for both