The Student Room Group

Sketching curves and finding points of intersection of two graphs

I'm just doing some revision over the stuff I'm stuck on and it's the first question on the mixed exercise and I'm stuck on it. Unbelieveable.

1a. On the same axes sketch the graph sof y = x^2 (x - 2) and y = 2x - x^2
b. By solving a suitable equation find the points of intersection of the two graphs

Ok, sketching the curve is: y = x^2 (x-2) - a postive cubic curve touching the x axis.
etc etc

What I'm stuck on is finding the points of intersection.
I put it to "y=y"
x^2 (x - 2) = 2x - x^2
x^3 - 2x^2 = 2x - x^2
x^3 - x^2 - 2x = 0
x ( -x^2 - x - 2 ) = 0
x [ (-2 - x) (1 + x) ]
x = -2 x = -1

now I dont know what to do. Also theres going to be three intersections I've only got 2 x values there =/

Reply 1

ShaolinTemple
I'm just doing some revision over the stuff I'm stuck on and it's the first question on the mixed exercise and I'm stuck on it. Unbelieveable.

1a. On the same axes sketch the graph sof y = x^2 (x - 2) and y = 2x - x^2
b. By solving a suitable equation find the points of intersection of the two graphs

Ok, sketching the curve is: y = x^2 (x-2) - a postive cubic curve touching the x axis.
etc etc

What I'm stuck on is finding the points of intersection.
I put it to "y=y"
x^2 (x - 2) = 2x - x^2
x^3 - 2x^2 = 2x - x^2
x^3 - x^2 - 2x = 0
x ( x^2 - x - 2 ) = 0
x [ (-2 - x) (1 + x) ]
x = -2 x = -1

now I dont know what to do. Also theres going to be three intersections I've only got 2 x values there =/

Is x=0 not also a solution? :p:

Reply 2

missing x=0

edit: Billy, you're too quick!

Reply 3

actually what am I doing


x ( -x^2 - x - 2 ) = 0
doesn't even factorise

what should I do then

Reply 4

It's x(x^2 - x - 2) = 0 not x(-x^2 -x - 2) = 0.

Reply 5

sigh *slap self*

Ok so x = 0, -1, 2
the coordinates are therefore (0,0) (2,0) (-1,-3)
How do you get the (-1,-3)? do you put the x value of -1 and 2 into one of the equations?

Reply 6

Oh I got it now, thanks

I can move onto question 2 now, which I'm going to be stuck on straight away

Reply 7

Time for question 2!

Sketch the curves y = 6/x y= 1+x

find the coordinates at the points of intersection A and B.
6/x = 1+x
6 = 1x + x^2
x^2 +1x - 6
(x + 3)(x - 2)
so the coordinates are
(-3,-2) (2,3)


part c.

The curve C with equation y = x^2 + px + q, where p and q are integers, passes through A and B. Find the values of p and q

Don't know where to start

Reply 8

Firstly, in your working, you seem to have lost the equal sign about half way through :p:

You know the quadratic passes through (-3,-2) and (2,3). What does that actually mean? It means that when x is -3, y is -2. When x is 2, y is 3. So how can you apply that to y = x^2 + px + q?

Reply 9

So you get simultaneous equations?

-1 = 2p + q
-11 = -3p +q

Subract them to get
10 = 5p
p = 2

y = x^2 + 2x + q

Reply 10

Yeah. Now just solve them.