# Im so stuck point on graph

• Jan 15th 2007, 07:35 AM
jojo
Im so stuck point on graph
I think this is the right forum, probably wrong but oh well :)

f(x)=ax3-3x. The point (-2,0) lies on the graph of y=f(x). Find the value of a.

Thanks for the future help ppl.

xxJoJoxx
• Jan 15th 2007, 07:47 AM
CaptainBlack
Quote:

Originally Posted by jojo
I think this is the right forum, probably wrong but oh well :)

f(x)=ax3-3x. The point (-2,0) lies on the graph of y=f(x). Find the value of a.

Thanks for the future help ppl.

xxJoJoxx

The point (-2,0) lies on y-f(x), so f(-2)=0, so:

a(-2)^3-3(-2)=-8a+6=0,

so rearranging:

8a=6,

or a=3/4.

RonL
• Jan 15th 2007, 07:54 AM
jojo
Hello again :D soz to ask but theres a b bit :) I swear I'm actually good at maths I've been off for 2 weeks tho :(

"Find the coordinates of the point where the line 8y-6x=48 cuts the X-axis."

I think its that x=0 thing but cant remember. Thanks again
• Jan 15th 2007, 07:57 AM
CaptainBlack
Quote:

Originally Posted by jojo
Hello again :D soz to ask but theres a b bit :) I swear I'm actually good at maths I've been off for 2 weeks tho :(

"Find the coordinates of the point where the line 8y-6x=48 cuts the X-axis."

I think its that x=0 thing but cant remember. Thanks again

No it cuts the x-axis when y=0. So you need to solve:

-6x=48,

which gives x=-8.

RonL