#1

I'm doing AS at college and I'm stuck on this question.

There's four different ones but I need to know the technique, can someone give me something step by step or a link?

Question: Find the points of intersection of...

a. y=x^2+3x-7

y=3

There's four different ones but I need to know the technique, can someone give me something step by step or a link?

Question: Find the points of intersection of...

a. y=x^2+3x-7

y=3

#2

The point of intersection is when one graph = the other.

So y = y

x^2 +3x - 7 = 3

x^2 + 3x - 10 = 0

Then solve as a quadratic

(x + 5)(x - 2) = 0

x = -5 or x = 2

You know that the y coordinate will be 3, so;

(-5, 3) and (2, 3)

So y = y

x^2 +3x - 7 = 3

x^2 + 3x - 10 = 0

Then solve as a quadratic

(x + 5)(x - 2) = 0

x = -5 or x = 2

You know that the y coordinate will be 3, so;

(-5, 3) and (2, 3)

#3

***Brain Explodes***

#4

Wow, cheers, that looks pretty easy.

One last one:

Find the value of C for which y=x+c is a tangent to y=3-x-5x^2

One last one:

Find the value of C for which y=x+c is a tangent to y=3-x-5x^2

#5

The point of intersection is when one graph = the other.

So y = y

x^2 +3x - 7 = 3

x^2 + 3x - 10 = 0

Then solve as a quadratic

(x + 5)(x - 2) = 0

x = -5 or x = 2

You know that the y coordinate will be 3, so;

(-5, 3) and (2, 3)

Perfect

#6

Wow, cheers, that looks pretty easy.

One last one:

Find the value of C for which y=x+c is a tangent to y=3-x-5x^2

I'm going to work this out as though you do;

The gradient of the tangent is the same as the gradient of the line at any given point.

When x = 0, y = 3, so the line goes through (0,3)

The gradient of the line is dy/dx

dy/dx = -1 - 10x

At the point (0,3) x = 0, so

dy/dx = m = -1

y = mx + c

y = -1x + c

At x = 0

3 = 0 + c

Hence, y = mx + c

y = -1x + 3

c = 3

That's all I can think to do with that, if you didn't mean y = mx + c, I'm not entirely sure what to do. If you get back to me and let me know whether or not it was y = mx + c I can have a go.

EDIT: Hold on. I'll re-edit in a minute.

Okay, at y = x + c, m = 1 (y = mx + c).

Therefore dy/dx of the line = 1.

-1 - 10x = 1

-10 x = 2

x = -(1/5)

at -(1/5)

y = 3 - x - 5x^2

y = 3 + 1/5 - 1/5

(5x^2 = (1/25) x 5 = 5/25 = 1/5)

y = 3

Back to the tangent;

y = x + c

3 = -(1/5) + c

c = 3 + 1/5

c = 3.2

Though this could be wrong, not sure. Check the back of your book.

*Last edited by JamieB at Nov 18, 2007,*

#7

It was 16 over 5.

#8

Wow, cheers, that looks pretty easy.

One last one:

Find the value of C for which y=x+c is a tangent to y=3-x-5x^2

Ok, if you meant it as you wrote it, then you have to the point on the curve that has gradient=1

So: dy/dx=-1-10x=1 => -10x=2 => x=-1/5

now, at x=1/5, y=3+1/5-5/25=3

The tangent must now touch the curve at(-1/5,3), so we heve the constraint for c that 3=-1/5+c =>c=3+1/5 = 16/5

#9

Ok, if you meant it as you wrote it, then you have to the point on the curve that has gradient=1

So: dy/dx=-1-10x=1 => 10x=2 => x=1/5

now, at x=1/5, y=3-1/5-5/25=3

The tangent must now touch the curve at(1/5,3), so we heve the constraint for c that 3=1/5+c =>c=3-1/5 = 14/5

It's 16 over 5. Anywhere you could of gone wrong?

And what does the D mean?

#10

It was 16 over 5.

16/5 = 3.2

#11

Well i messed that up...

Nothing to see here.

Nothing to see here.

#12

It's 16 over 5. Anywhere you could of gone wrong?

And what does the D mean?

Yeah i know, i fixed that . What D are youtalking about?

#13

Yeah i know, i fixed that . What D are youtalking about?

dx/dy D.

EDIT: Also, in reference to the first question, what do you do when you don't have a explict Y axis? ie...

y=5x^2-2x+1

y=6-3x-x^2

#14

dx/dy D.

It means delta. Basically, the increment of x over 1y divided by the increment of y over 1x.

#15

It means delta. Basically, the increment of x over 1y divided by the increment of y over 1x.

Rephrase using a practical example.

#16

This is my favorite kind of math

edit: but i can't do the second question due to it having a minus sign.

edit: but i can't do the second question due to it having a minus sign.

#17

Set them equal to eachother. Is that C1?

#18

d means delta which means "change in".

So dy/dx means how much the line has gone up or down per 1x

The gradient.

Im a ****e explainer.

So dy/dx means how much the line has gone up or down per 1x

The gradient.

Im a ****e explainer.

#19

Set them equal to eachother. Is that C1?

Bingo.

How'd ya approach that?

#20

It was 16 over 5.

16/5 = 3.2

So look at my working after the edit, it's right.

#21

Rephrase using a practical example.

It's the change in X and Y.

First X - the Second X divided by the first Y minus the second Y, doing your subtracting before dividing (pretend there's brackets).

#22

d means delta which means "change in".

So dy/dx means how much the line has gone up or down per 1x

The gradient.

Im a ****e explainer.

Got it.

And the gradient = y/x.

Never seen delta before, thats all.

#23

Do you mean y = mx + c?

I'm going to work this out as though you do;

The gradient of the tangent is the same as the gradient of the line at any given point.

When x = 0, y = 3, so the line goes through (0,3)

The gradient of the line is dy/dx

dy/dx = -1 - 10x

At the point (0,3) x = 0, so

dy/dx = m = -1

y = mx + c

y = -1x + c

At x = 0

3 = 0 + c

Hence, y = mx + c

y = -1x + 3

c = 3

That's all I can think to do with that, if you didn't mean y = mx + c, I'm not entirely sure what to do. If you get back to me and let me know whether or not it was y = mx + c I can have a go.

EDIT: Hold on. I'll re-edit in a minute.

Okay, at y = x + c, m = 1 (y = mx + c).

Therefore dy/dx of the line = 1.

-1 - 10x = 1

-10 x = 2

x = -(1/5)

at -(1/5)

y = 3 - x - 5x^2

y = 3 + 1/5 - 1/5

(5x^2 = (1/25) x 5 = 5/25 = 1/5)

y = 3

Back to the tangent;

y = x + c

3 = -(1/5) + c

c = 3 + 1/5

c = 3.2

Though this could be wrong, not sure. Check the back of your book.

This (the 2nd half) is right

#24

It's the change in X and Y.

First X - the Second X divided by the first Y minus the second Y, doing your subtracting before dividing (pretend there's brackets).

Other way around lol.

(Y1 - Y2) / (X1 - X2)

#25

dx/dy D.

EDIT: Also, in reference to the first question, what do you do when you don't have a explict Y axis? ie...

y=5x^2-2x+1

y=6-3x-x^2

dy/dx is a notation used by Liebnitz (co-founder in his own right of calculus) used to describe an infinitessimal increase in the y-direction divided by an infinitessimal increase in the x-direction. This is what gives you the gradient of a point, which you should be able to realise seems nonsensical at first (how can something 0-dimensional have a gradient?). Please note its not a d like the 4th letter of the alphabet d, it comes from the greek letter delta which is used in the sciences to describe a change of some description.

What do you mean by explicit y-axis?

#26

dy/dx is a notation used by Liebnitz (co-founder in his own right of calculus) used to describe an infinitessimal increase in the y-direction divided by an infinitessimal increase in the x-direction. This is what gives you the gradient of a point, which you should be able to realise seems nonsensical at first (how can something 0-dimensional have a gradient?). Please note its not a d like the 4th letter of the alphabet d, it comes from the greek letter delta which is used in the sciences to describe a change of some description.

What do you mean by explicit y-axis?

In the previous example, we had a y=3. In this, we have an expression.

#27

In the previous example, we had a y=3. In this, we have an expression.

ah, well you do the same thing.

if you have y=x^2-6x+4 and y=7x^2-9 you can set them equal to one another, since x^2-6x+4=y=7x^2-9, hence 7x^2-9=x^2-6x+4.

#28

dx/dy D.

EDIT: Also, in reference to the first question, what do you do when you don't have a explict Y axis? ie...

y=5x^2-2x+1

y=6-3x-x^2

Same as before

5x^2 - 2x + 1 = 6 - 3x - x^2

6x^2 + x - 5 = 0

(6x - 5)(x + 1)

x = 5/6 or x = -1

#29

i'm a wee bit stuck on a stats question so I thought i'd post it here instead of creating a new thread

the question (exactly how it is in the book) is

The r.v. X is such that X ~ N(μ, 4). A random sample, size n , is taken from the population.

Find the least n such that P(| - μ| < 0.5) >0.95

the question (exactly how it is in the book) is

The r.v. X is such that X ~ N(μ, 4). A random sample, size n , is taken from the population.

Find the least n such that P(| - μ| < 0.5) >0.95

#30

Same as before

5x^2 - 2x + 1 = 6 - 3x - x^2

6x^2 + x - 5 = 0

(6x - 5)(x + 1)

x = 5/6 or x = -1

Y being...

#31

Y being...

Plug the X values into the original equations and find out

#32

Plug the X values into the original equations and find out

Ohh yeah, sorry, haha.

I just solved one before, completely forgot.