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# Maths Question?

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1. If the line y=3 is a tangent to a curve, does the tangent have a gradient of infinity or 0?
2. (Original post by mil88)
If the line y=3 is a tangent to a curve, does the tangent have a gradient of infinity or 0?
Zero - the line is horizontal.
3. (Original post by joostan)
Zero - the line is horizontal.
Would it be different for x=3?
4. (Original post by mil88)
If the line y=3 is a tangent to a curve, does the tangent have a gradient of infinity or 0?
Zero, as the y co-ordinates aren't increasing by anything the further you go along.

This maybe relevant to the question you're doing depending on what it is, but if you know that the gradient of a curve at a point (or dy/dx) is zero, what does this mean about the point?
5. (Original post by KaylaB)
Zero, as the y co-ordinates aren't increasing by anything the further you go along.

This maybe relevant to the question you're doing depending on what it is, but if you know that the gradient of a curve at a point (or dy/dx) is zero, what does this mean about the point?
Note: I actually meant x=3

Well if the gradient is 0, then the point is a stationary point (if that's what you're getting at?)
6. (Original post by mil88)
Note: I actually meant x=3

Well if the gradient is 0, then the point is a stationary point (if that's what you're getting at?)
Yeah exactly, and the same principle applies for x=3 don't worry
7. (Original post by KaylaB)
Yeah exactly, and the same principle applies for x=3 don't worry
But why isn't it infinity for x=3, because surely the y values are increasing but x values aren't. So if you do, dy/dx: you would get a constant/0, which is infinity isn't?
8. (Original post by mil88)
But why isn't it infinity for x=3, because surely the y values are increasing but x values aren't. So if you do, dy/dx: you would get a constant/0, which is infinity isn't?
It just means no matter the y value, x will always be 3

Also, I'm not too sure what you meant in the bit I bolded
9. (Original post by mil88)
But why isn't it infinity for x=3, because surely the y values are increasing but x values aren't. So if you do, dy/dx: you would get a constant/0, which is infinity isn't?
there is no change in y/x values to get a dy/dx

it would be 0/constant= 0

(Oops thought you meant for y=3)
10. (Original post by KaylaB)
It just means no matter the y value, x will always be 3

Also, I'm not too sure what you meant in the bit I bolded
I meant that you will have a constant divided by 0.
11. All these complicated answers, I'll tell you, it's 69 🎉
12. (Original post by mil88)
Would it be different for x=3?
Well here it would be best to say that the gradient is undefined.
It is often convenient to think of this as being infinite, and you won't go too far wrong thinking of it this way.
13. (Original post by Dinaa)
there is no change in y/x values to get a dy/dx
Well using change in y divided by change in x, there will be a change in y and none in x, hence constant divided by 0 which is infinity isn't it?
14. (Original post by joostan)
Well here it would be best to say that the gradient is undefined.
So infinite right?
15. (Original post by mil88)
So infinite right?
Check my edit.
16. (Original post by KaylaB)
It just means no matter the y value, x will always be 3

Also, I'm not too sure what you meant in the bit I bolded
I think OP is talking about the change in y would be 0 and the change in x= a constant

therefore change in y/change in x= constant/0
17. (Original post by mil88)
Well using change in y divided by change in x, there will be a change in y and none in x, hence constant divided by 0 which is infinity isn't it?
yh I edited, thought u mean y=3 :3
18. (Original post by Dinaa)
I think OP is talking about the change in y would be 0 and the change in x= a constant

therefore change in y/change in x= constant/0
Ah okay thanks for clearing that up
19. (Original post by joostan)
Check my edit.
Ok thanks then! There was a question which I think required you to 'think' that gradient was infinite.
20. (Original post by Dinaa)
yh I edited, thought u mean y=3 :3
Nah don't worry, that's my fault for writing it incorrectly in my initial post. Thanks anyway.

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