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A Level Maths

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Reply 40

me!

JamesF

On the grid is there already a curve? If so, what is the equation of the curve?

there is - the one I drew for the previous question, it's equation is:

y = x^3 + 2

it's a bendy curve thing

Reply 41

me!

2776

Draw a line when x=0, and read off the figures for the solutions (ie when they cross the lines)

i don't understand what you mean :confused:

*thinks*

Reply 42

me!

i don't understand what you mean :confused:

*thinks*

Scratch that, I thought you were given the curve and only needed to draw the line x=0. Very naieve of me.

Reply 43

username9816

me!

You could help me if you want :smile:

"By drawing a suitable straight line on the grid find estimates of the solutions of the equation x^3 - 4x + 2 = 0"

It's from a past gcse paper - it confused me because it's not a straight line...

Plot some points to sketch the curve, then draw a horizontal line through y = 0, and take/mark off the values where the curve crosses this line that you drew.

Remember, the line y = 0 [b[is the x-axis (evidently), therefore the solutions to the equation x^3 - 4x + 2 = 0 is when the graph of this function cuts y = 0 i.e.) the x-axis.

Say the graph, for example, crosses the line y = 0 (the x-axis) at x = 2 and x = 4. The solutions to the equation x^3 - 4x + 2 = 0 would be x = 2, x = 4.

EDIT: IN LIGHT OF THE FACT THAT THERE IS INFACT ANOTHER GRAPH INVOLVED, AND NOT JUST THIS ONE, SCRAP WHAT I SAID ABOVE!

Reply 44

username9816

2776

Scratch that, I thought you were given the curve and only needed to draw the line x=0. Very naieve of me.

Ooops, I explained it like that as well, I didn't know she had another curve.....

Reply 45

username9816

2776

Scratch that, I thought you were given the curve and only needed to draw the line x=0. Very naieve of me.

You mean the line y = 0

Reply 46

JamesF

You have the line y = x^3 + 2, and you want the line y = x^3 - 4x + 2, so draw the line y = 4x
Now we have
4x = x^3 + 2
0 = x^3 - 4x + 2

So, when you draw the line y=4x, the solutions are the points where the line and curve cross.

Reply 47

username9816

2776

Scratch that, I thought you were given the curve and only needed to draw the line x=0. Very naieve of me.

The line y = 0 is the x-axis.
The line x = 0 is the y-axis.

Reply 48

me!

JamesF

cool - thanks :biggrin:

the only other questions annoying me on the paper were (5p^3 / q)^3
for which I got 125p^9/q^3, is this right?

and (12t^5/u^4)(u^3/3t^2) for which I got 4t^3/u...

Reply 49

JamesF

me!

cool - thanks :biggrin:

the only other questions annoying me on the paper were (5p^3 / q)^3
for which I got 125p^9/q^3, is this right?

and (12t^5/u^4)(u^3/3t^2) for which I got 4t^3/u...

Both correct :smile:

Reply 50

me!

JamesF

Both correct :smile:

wohoo - and thanks, if I get stuck again I know where to come

Reply 51

Dill

I do maths with stats, so its 2 modules of pure, and 1 of stats, and every week we have 200 mins pure, and 100 mins stats. if we had anything less than that it wouldn't be too good! we are rushing in pure as it is! 2 hours for maths&further maths is just plain ridiculous! that just doesn't seem fair to me!