# maths question

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

f(x) = 2x^2 - x

g(x) = x^3

(iii) solve the equation f(x) = g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer says (0,0) so I'm confused

g(x) = x^3

(iii) solve the equation f(x) = g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer says (0,0) so I'm confused

0

reply

Report

#2

(Original post by

f(x) = 2x^2 - x

g(x) = x^3

(iii) solve the equation f(x) = g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer says (0,0) so I'm confused

**Mia 15**)f(x) = 2x^2 - x

g(x) = x^3

(iii) solve the equation f(x) = g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer says (0,0) so I'm confused

(0,0) and (1,1)

sub in x=0, 1 into one of the equations

Last edited by Harribos; 1 month ago

1

reply

Report

#3

**Mia 15**)

f(x) = 2x^2 - x

g(x) = x^3

(iii) solve the equation f(x) = g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer says (0,0) so I'm confused

**complete**question? Are there other restrictions on x?

0

reply

thanks I understand that part now but I was confused why the answer book just had (0,0) written for that question. because if the question says solve the equation f(x)=g(x) then shouldn't the answer just be the x values you found such as x=1 and x=0

0

reply

(Original post by

Can you upload a picture of the

**davros**)Can you upload a picture of the

**complete**question? Are there other restrictions on x?Q) Two functions are given as f(x) = 2x^2 - x and g(x) = x^3.

(i) draw the graph

(ii) how many points of intersection are there? i wrote 2

(iii)

[a] solve the equation f(x) =g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer just says (0,0) so I'm confused

[b] hence find the coordinates of the points where the two curves intersect

when x=1, y = 2(1)^2 = 1 so (1,1)

when x=0, y=2(0)^2-0 so (0,0)

but the answer only says (1,1)

is this just an error in the answer book?

0

reply

Report

#6

(Original post by

Sorry the picture won't upload but here's the question word for word:

Q) Two functions are given as f(x) = 2x^2 - x and g(x) = x^3.

(i) draw the graph

(ii) how many points of intersection are there? i wrote 2

(iii)

[a] solve the equation f(x) =g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer just says (0,0) so I'm confused

[b] hence find the coordinates of the points where the two curves intersect

when x=1, y = 2(1)^2 = 1 so (1,1)

when x=0, y=2(0)^2-0 so (0,0)

but the answer only says (1,1)

is this just an error in the answer book?

**Mia 15**)Sorry the picture won't upload but here's the question word for word:

Q) Two functions are given as f(x) = 2x^2 - x and g(x) = x^3.

(i) draw the graph

(ii) how many points of intersection are there? i wrote 2

(iii)

[a] solve the equation f(x) =g(x)

I did 2x^2 - x = x^3

-x^3 + 2x^2 - x = 0

x = 1 and x = 0

but the answer just says (0,0) so I'm confused

[b] hence find the coordinates of the points where the two curves intersect

when x=1, y = 2(1)^2 = 1 so (1,1)

when x=0, y=2(0)^2-0 so (0,0)

but the answer only says (1,1)

is this just an error in the answer book?

-x(x-1)^2

The curves graze at 1 (double root) but cross (intersect?) at 0

https://www.desmos.com/calculator/fckj5v9zbs

It would help to see the exact wording in the original question.

Last edited by mqb2766; 1 month ago

1

reply

(Original post by

You realise the combined curve factorizes as

-x(x-1)^2

The curves graze at 1 (double root) but cross (intersect?) at 0

https://www.desmos.com/calculator/fckj5v9zbs

It would help to see the exact wording in the original question.

**mqb2766**)You realise the combined curve factorizes as

-x(x-1)^2

The curves graze at 1 (double root) but cross (intersect?) at 0

https://www.desmos.com/calculator/fckj5v9zbs

It would help to see the exact wording in the original question.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top