Question

I'm stuck, but solved most of it!

Answer 1

Okay, so after a lot of calculating, I found that m= -3; m= -2. My problem is lies in putting it in the equation form.

Answer 2

You have a point and you have a slope. So write it in point slope form?

Answer 3

\[y- y_1 = m(x - x_1)\]

Answer 4

So I should do it twice then, once for each slope? :p

Answer 5

It seems like it's asking for two lines, so yes.
(Disclaimer: I didn't actually do the problem, I'm just going by what you said)

Answer 6

Alright, I got-3x+7 and -2x+5...

Answer 7

I hope they are right :p

Answer 8

Damn it...I miscalculated somewhere I guess T-T

Answer 9

Do you want me to try to go through the whole problem with you?

Answer 10

I would certainly appreciate it! :0

Answer 11

Ok, so, our parabola is \(x^2 + x + 4\)

Answer 12

I assume you know how to take the derivative of that

Answer 13

2x+1

Answer 14

Right, so that's the slope of a tangent line

Answer 15

Slope, right.

Answer 16

Two points make a line. One point is given. The other point has to be on the curve. Let's call it z. So the coordinates there are \((z, z^2 + z + 4)\)

Answer 17

With me so far?

Answer 18

I am! I'll respond so it doesn't seem I am away.

Answer 19

We'll use point slope form again to write our line. Gimme a sec to think so I don't mess it up.

Answer 20

Alrighty

Answer 21

\[y - (z^2 + z + 4) = (2z + 1)(x - z)\]

Answer 22

Do you understand this somewhat hideous expression?

Answer 23

If you substitute in the given point, that is, x = 2 and y = 1, and solve for z, you'll end up with a quadratic equation with two roots.

Answer 24

Yeah....I do...somewhat xP

Answer 25

All you're doing is finding the values of z that we gave before, i.e., the x-values on the curve where the line is tangent and goes through the given point.

Answer 26

Mmmm...one sec!

Answer 27

z= -sqrt(5)/4

Answer 28

I don't think that is the right answer...

Answer 29

You started with \(1 - (z^2 + z + 4) = (2z + 1)(2 - z)\) right?

Answer 30

I am doubting it to. I did! I was passing things all over the place and...well, got messy.

Answer 31

Ok, so we'll expand the right side, first. FOIL! \(-2z^2 + 3z + 2\)

Answer 32

Okay, I got that...

Answer 33

Now on the left, just make the whole thing negative and add 1. There's probably a "nicer" way to do it but it's easy to think about, I think.

\(1 + -(z^2 + z + 4) = -z^2 - z - 4 + 1\)

Answer 34

Okay, I understood. I was off with the signs on that part.

Answer 35

Then we can simplify the whole mess down to \( -z^2 + 4z + 5 = 0\)

Answer 36

Now we want to solve for z?

Answer 37

Yes. This one actually factors nicely but if you hate factoring like me just use the quadratic formula.

Answer 38

I am gonna go with the formula.

Answer 39

http://www.mathsisfun.com/quadratic-equation-solver.html

Answer 40

I am almost done. I wanna try myself :p

Answer 41

Go for it. It's handy when you're tired of crunching the quadratic formula for the 20th time though. :D

Answer 42

-1, 5.

Answer 43

It is really handy xP

Answer 44

Yes. So those values are the x-values along our curve where there is a tangent line.

Answer 45

Alright ._.

Answer 46

Do you need me to go on or do you get it? The rest is just more crunching.

Answer 47

How are your teeth?

Answer 48

:D <-- just fine!

Answer 49

Perfect for more crunching xD!

Answer 50

Ok. So we know the x-values. And we have the equation of the curve. So we can get y-values.

Answer 51

Let's do it.

Answer 52

\[(-1)^2 + (-1) + 4 = 4\]

Answer 53

\[(5)^2 + 5 + 4 = 34\]

Answer 54

Alright, so we should move on to y=mx+b or that other fancy one? :p

Answer 55

We don't know the y-intercept so we need the fancy one.

Answer 56

So we have two lines. One line between (2,1) and (-1,4), and line between (2,1) and (5,34)

Answer 57

So find the slopes.

Answer 58

And that will get us back to where we started, only with correct answers, hopefully. :D

Answer 59

Wait, so x-values are -1, 5 and y's are 2,1?

Answer 60

No. We have two lines.

Answer 61

Both lines have (2,1) as a point on that line, because that was the point the problem specified.

Answer 62

Then the other point on each line is based one one of the z-solutions we found before. And then plugged back in.

Answer 63

So the line for the -1 solution of z goes from (2,1) to (-1,4)

Answer 64

And the line for the 5 solution of z goes from (2,1) to (5,34)

Answer 65

-11, -3 T-T

Answer 66

You got the sign wrong. 2 - 5 = -3, 1 - 34 = -33, so -33/-3 = 11

Answer 67

God damn it.

Answer 68

And the other one... That's 3/-3, so -1

Answer 69

So the slopes are 11 and -1. Now you can scrooooll all the way up and do the solution I gave you when you knew the slopes. We are back to where we started, with the right answers!

Answer 70

So I am using the fancy one again?

Answer 71

Point slope form, yes.

Answer 72

Okay...I got y=11x-12 and -x-1...no reason for it to be wrong O.O

Answer 73

\[y-1 = -1(x - 2)\]\[y = -x + 2 + 1\]\[y = -x + 3\]

Answer 74

I'll let you try the other one again. :D

Answer 75

Wait, I put a -2 because the formula already includes a negative sign, I figured -(-2), no?

Answer 76

But the point is (2,1). So it's (x - 2)

Answer 77

Yes, it's true...poop.

Answer 78

Is the other one wrong?

Answer 79

\[y - 1 = 11(x - 2)\]\[y = 11x - 22 + 1\]\[y =11x -21\]

So yes. You swapped the digits or something.

Answer 80

I am going to input the answers then. Wish you luck...

Answer 81

In order: -x+3 then other?

Answer 82

Yes, that's the smaller one.

Answer 83

At least I got something right.

Answer 84

You are right. I can't even begin to thank you. Take my medal, fan and all my gratitude. You are a saint.

Answer 85

I'm glad it ended up being right! I would've felt horrible if it ended up wrong after all that!

Answer 86

Speak Spanish?

Answer 87

Si, un poco

Answer 88

Ah, ya veo. Bueno, tenga una buena noche. And again, thank you.

Answer 89

You too!