x Turn on thread page Beta
 You are Here: Home >< Maths

Edexcel C1 finding x coordinates graph curve watch

1. I would really appreciate any help here. Thanks!!

The curve C has equation
y= (x+3) (x-1)^2
a. sketch C showing clearly coordinates of points where curve meets coordinate axes.
I got points (-3,0) (1,0) (1,0) (0,3) (0,-1) are they right? If so, then how come the graph won't turn out properly if I try to connect them?? If they're not right, what are the right points?
b. Show that equation of C can be written in the form
y= x^3 + x^2-5x+k
where k is a positive integer, and state the value of k.
Got this right. k =3

There are two points on C where gradient of the tangent to C is equal to 3.
c. Find x coordinates of these two points.
Help here?
2. (Original post by BrownieLover)
I would really appreciate any help here. Thanks!!

The curve C has equation
y= (x+3) (x-1)^2
a. sketch C showing clearly coordinates of points where curve meets coordinate axes.
I got points (-3,0) (1,0) (1,0) (0,3) (0,-1) are they right? If so, then how come the graph won't turn out properly if I try to connect them?? If they're not right, what are the right points?
b. Show that equation of C can be written in the form
y= x^3 + x^2-5x+k
where k is a positive integer, and state the value of k.
Got this right. k =3

There are two points on C where gradient of the tangent to C is equal to 3.
c. Find x coordinates of these two points.
Help here?
Well this is a cubic function with a repeated root, hence it crosses the x-axis just twice. - (-3,0) and (1,0). The y-intercept is found by subbing X=0.. Which is (0,3).

How do you find the gradient of a polynomial ?
3. (Original post by MathsLover28)
Well this is a cubic function with a repeated root, hence it crosses the x-axis just twice. - (-3,0) and (1,0). The y-intercept is found by subbing X=0.. Which is (0,3).

How do you find the gradient of a polynomial ?

yeah but when we sub x=0 for (x-1) we're also gonna get (0, -1) so there are two points on the y axis aren't there?
4. (Original post by BrownieLover)
yeah but when we sub x=0 for (x-1) we're also gonna get (0, -1) so there are two points on the y axis aren't there?
Nooo....

let's change the notation around a bit.

so what is f(0), well where ever there's an x we put 0, so :

The product of two brackets, implies there are two roots (when the function crosses the x-axis), not the y-intercept. Y is a function of x, not the other way around.
5. (Original post by BrownieLover)
yeah but when we sub x=0 for (x-1) we're also gonna get (0, -1) so there are two points on the y axis aren't there?
How can you get two values for y when you substitute one value for x? That's not a function, is it?

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 31, 2015
Today on TSR

Loughborough better than Cambridge

Loughborough at number one

Poll
Useful resources

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Chat with other maths applicants