Drawing graphs help!

Hi everyone!

Basically I could be popping up quite often in the next few days because C3 coming up and all that jazz and I only just got round to opening my maths book...add to that the fact that maths is not exactly my strong point and one winning combination for annoying questions that you all probably find very simple but here goes...

...this is OCR by the way.

Ok, I'm stuck on a question in which the first part says:

It is is given that a and b are positive constants. By sketching graphs of:

y = x^5

and y = a - bx

on the same diagram, show that the equation

y = x^5 + bx + a = 0

has exactly one real root.

Now, I am entirely stuck so I have nothing to show for attempting it - I have the mark scheme but I don't understand what it says.

Can anyone tell me, what topic is this? How do I sketch the

y = x^5

graph? Am I meant to know how for C3? Can anyone give me any pointers that might help me to make sense of how to do the question and what topic it's from?

Thanks to everyone who helps!!

Reply 1

generalebriety

You're meant to know how to sketch y = x^n for positive integers n just by 'guessing' from the graphs of y = x^2 and y = x^3. Essentially, y = x^(odd) will "look like" y = x^3 and y = x^(even) will "look like" y = x^2. Also, y = (some polynomial of degree n) can cross the x-axis at most n times, so can have at most (n-1) turning points. That's all you need to know.

It should be obvious that if you draw y = x^5 and y = -a-bx, and find where they intersect, you'll have solved the equation x^5 = -a - bx, right? But that's the equation you wanted to know stuff about...

Reply 2

hunibuni

Sorry....this is in fact the 2nd half of the same question but you'll be glad to know I have sort of done it this time and I have the right answer according to the markscheme but would someone be able to check my working for me? Thanks!

Q: Use the iterative formula

x_n+1

\sqrt[5]{53-2x_n}

with a suitable starting value to find the real root of the equation

x^5 + 2x - 53=0

. Show the result of each iteration, and give the root correct to 3 dp.

where x = 2, equation gives -17 = too small
where x = 3, equation gives 196 = too big

Therefore let

x_1 = 2.5

x_2 = \sqrt[5]{53-5} = 2.168943542[br]x_3 = \sqrt[5]{53-ansx2} = 2.174894487[br]x_4 = \sqrt[5]{53-ansx2} = 2.174788089

root = 2.175 (3.d.p)

Reply 3

generalebriety

You'd probably lose a mark or two for that - just because it seems to be settling at about 2.175, you can't assume it is. You have to work out f(2.1745) and f(2.1755) (the upper and lower bounds for '2.175 to 3 d.p.') and show that there's a sign change, so the root is definitely in that interval, before you can conclude it.

Reply 4

Swayum

You should basically know how to sketch y = x^5 from C1. Here's how I sketch polynomial graphs:

First determine the roots of the equation y = x^5. I hope you can see it has one root, x = 0. So this means it will cross the x axis only once.

Next determine the y intercept. This occurs when x = 0 and so y = 0 as well.

So the graph cuts the x axis and y axis at (0,0) in other words the origin.

The final thing you need to consider is the shape of the graph. When you've got an even power (by even power, I mean the highest power of x), the graph starts high and ends high. For example:

y = x^2 + 4x - 9 looks like this

y = x^4 + 2x^3 - x - 8 looks like this

And the part of starting high and ending high holds for all even powers.

When you have odd powers, the graph starts low and ends high. For example:

y = x^3 + 4x^2 - 5x + 8 looks like this

Ok, so y = x^5 would start low and end high. And we know it passes through the origin. It would therefore look like this:

Now sketch that on a bit of paper and sketch y = a - bx on the same graph. Where the graphs intersect, their equations would be the same. So

x^5 = a - bx

Can you see how to do the question now?

Reply 5

hunibuni

oops...latex doesn't recognise line spaces I take it!! sorry!

Thank you sooooooo much generalebriety - that was so helpful!!

ok, so the y = x^5 graph will just be the curve through 0 like y = x^3?

Then the straight line, the equation means it will have a negative gradient and it cuts the y axis at a?

Reply 6

hunibuni

Oh ok....sorry I hadn't seen your post swayam!

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