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
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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
Hello again 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
Originally Posted by jojo Hello again 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
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