Question

Please Help!

The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes through the point (0,-8) which is not on the graph of f. Find the values of a and b.

Answer 1

f(x)= x^3 - x^2 - 4x + 4

Answer 2

First of all, remember that the derivative of that equation is the line tangent to it.

Answer 3

f '(x)=3x^2 -2x - 4

Answer 4

now what do I do?

Answer 5

Hmm, I guess derive again to get the tangent line, since that's still a curve.

Answer 6

Then find a line that's consistent with (a, b) given.

Answer 7

so find the slope of the tangent line in terms of a and b and plug in 0 and -8?

Answer 8

hold on, doing some research. I think I may have led you astray - I think the right thing is probably to just take the first derivative and then plug that in as the slope of a line in point-slope form. Where point is (x, f(x)), and the slope is f'(x).

Answer 9

y-y1=m(x-x1)

Answer 10

right, where x1 is x, y1 is f(x), and m is f'(x)

Answer 11

no x1 and y1 is the point

Answer 12

To keep the x's and y's from getting mixed up, use a different x and y for the point-slope equation -- e.g. xx and yy

yy- f(x) = f'(x)(xx - x)

Answer 13

Then plug in the point (0, -8) you're given as xx and yy, and solve for x.

Answer 14

So it would be

-8 - f(x) = f'(x)(0 - x), and solve for x.

Answer 15

I think that should do it.

Answer 16

did that make sense?

Answer 17

i think that is wrong for some reason, I have never seen the point slope formula been used like that before

Answer 18

Neither had I until I looked up how to find the equation of a tangent line to a curve on wikipedia just now

Answer 19

the answer is a=2 and b=0

Answer 20

thats not what I necessarily got, but thats what her answer key says

Answer 21

hold on, let me type my equation into mathematica and see if that's the answer it gives

Answer 22

yep, it solves to 3 solutions, one of which is real and is x=2

Answer 23

and f(2) = 0

Answer 24

i still feel like there was a better way of doing this

Answer 25

cause I'm sell confused

Answer 26

still*

Answer 27

well, to define any line, you need two things: a point on the line, and its slope.

Answer 28

right

Answer 29

At any point x on your curve, the tangent line is a line that passes through the curve -- i.e., the point (x, f(x)), and has a slope f'(x). Does that make sense?

Answer 30

To define a tangent line at x, we need a point and a slope. The slope by definition is f'(x), and a point is (x, f(x)).

Answer 31

yes the slope of the tan line is the deriv. of f(x) and the point is usually either given or just the x value and you plug in the x value to f(x) to get the y value too

Answer 32

Right, the slope is the deriv of f(x),and we know one point is (x, f(x)). we have to find the value of X such that the line also passes through (0, -8). so the way I found it is used the point-slope form of the line to define the line, then solved it to see which of those tangent lines is also consistent with a=0, b=-8.

Answer 33

its like backwards from what I usually do, no?

Answer 34

they didn't give the "x" value
they just said it was a?

Answer 35

Ah - they used a, b, that's better than xx, yy. So the point-slope form of a line, if we're using a and b instead of x and y, is:

(b - y1) = m(a-x1)

We know a point on the tangent line is (x, f(x)), that's your x1 and y1. We know another point is (0, 8) so you plug that in for a, b. We also know m = f'(x), so you plug that in. You get a big equation that you solve for x, and when you do it solves to x=2. (and 2 complex numbers)

Answer 36

k stay here imma solve that

Answer 37

k

Answer 38

wait it's 0, -8 not 0,8, sorry, that was a typo.

Answer 39

i know i gotcha

Answer 40

(-12 + 2x^3 - x^2)=0

Answer 41

:/

Answer 42

imma get three answers for that, and I'm terrible at factoring...

Answer 43

well you can factor x^2 out of the last 2 terms and get the real 0

Answer 44

(-12 + 2x^3 - x^2) = (-2 + x) (6 + 3 x + 2 x^2)

Answer 45

-12 + (x^2)(2x-1)=0

Answer 46

i guess with the factoring above you get one of the zeroes just by inspection at x=2, and the other two by using the quadratic equation on the second term

Answer 47

sorry ya lost me

Answer 48

should i do p?'s over q?'s ???

Answer 49

well if you get to this:
(-12 + 2x^3 - x^2) = (-2 + x) (6 + 3 x + 2 x^2) = 0

.. then you know it's equal to 0 if either -2+x = 0,
or x^2 + 3x + 6 = 0,.

so the -2 + x = 0 you can solve easily, and the other two roots you can get by plugging into the quadratic equation.

Answer 50

no see
i now how to solve from x after you factor, but, how do you get from
(-12 + 2x^3 - x^2)
to
(-2 + x) (6 + 3 x + 2 x^2) = 0

Answer 51

without like a calc

Answer 52

haha, i don't actually know, i suck at factoring too to be honest. I used a calculator.

Answer 53

>_<

Answer 54

i can't use one UGH

Answer 55

my final is tomorrow and friday

Answer 56

and this is part c of question 1 of my free response

Answer 57

this is a practice question or your actual exam?

Answer 58

this is one of the practice questions from our AP CALC AB review packet problems

Answer 59

tomorrow is the "free response" section

Answer 60

ah. hmm... well, let me try and find an online factoring tutorial ;-)

Answer 61

so I'm doing the free response questions in our packet

Answer 62

lol factoring isnt even what our thing is on

Answer 63

its like limits, derivatives, optimization, theorems, related rates

Answer 64

yeah, pretty silly that's where we're stuck

Answer 65

I forgot factoring 20 years ago because the computer usually just does it for you anyway

Answer 66

I hate high school, I can't wait to get the heck out of here and move on to college

Answer 67

yeah, math education doesn't get much better there. too much focus on the mechanics (stupid stuff like factoring) and not enough focus on how to model problems (the interesting part of what we did on this problem)

Answer 68

well hopefully i will be able to test out of it with a good AP score this may

Answer 69

not to discourage you (you sound like a smart guy or girl) but, i tested out of early calc with my calc AB AP exam, and found every other math class I took was the same, way too focused on how to do the computations and not enough on understanding the concepts

Answer 70

I didn't really start to enjoy math until I started using programs like mathematica and ocatave that would do all the computations for me, freeing me to actually *understand* the math

Answer 71

not that this helps you for tomorrow

Answer 72

well i will be studying business so my math stuff will be in a bit different field, but thats okay i love all forms of math (WHEN IM NOT PRESSURED and I GET TIME TO LEARN), and i have been on my hs's math team for 4 years.

Answer 73

cool! being good at math will set you apart in business. what college?

Answer 74

can i tell u in a private chat?

Answer 75

is that possible on here??

Answer 76

no .. that's ok, not important

Answer 77

well to make a long story short

Answer 78

the common app's don't tell u till like xmas time so thats coming up soon, but to two other schools I have applied to that were not common app, I got in to, with money, direct admittance to their business schools and their Honors business programs, those schools being Indiana U and the U. of Texas. My common apps are Boston College, Villanova, Notre Dame, U. So. Cali (USC), U. of San Diego, Santa Clara U, Maryland U,

Answer 79

and Upenn (wharton)

Answer 80

wow, all excellent schools

Answer 81

haha ya very scattered but I'm intentionally trying to go as far away from home as possible

Answer 82

haha, you dont like home?

Answer 83

do you know what stir crazy is?

Answer 84

that's me....everyday

Answer 85

yep :)

Answer 86

what part of the US are you in?

Answer 87

Chicago is one of the best cities in the world, but I need to branch out. I seriously want to see the world out there, and it seems silly to just stay in one part of the world for most of your life.

Answer 88

totally agree. travel makes people 10x more interesting, 10x more understanding of cultures not their own, 10x more empathic

Answer 89

the least tolerant people I know invariably are the least traveled

Answer 90

I did this camp at georgetown 2 summers ago for 5 weeks. i meet people from everywhere in the world, in particular, 6 girls from Puerto Rico, who turned out to be my best friends i will ever have. I visited them last winter and it was just so cool. Im a guy by the way, so when I came back to school most of my friends where shocked to see me hanging out with all these people, especially the good looking P.R. girls haha

Answer 91

but if i never ventured, I would of never met them, and that would be a shame

Answer 92

haha, nice

Answer 93

and camp seems lame, it was a college enrichment program, u take 2 college courses and for fun do w/e u want around DC

Answer 94

that's where i grew up

Answer 95

Gtown is very nice!

Answer 96

lol loved waking up everyday near M street

Answer 97

yep, i lived in dupont circle for a couple of years, baltimore for 5 (in college), then israel, then LA, then seattle

Answer 98

wow all over haha thats the way !