Question

the graph of y=bx^2-cx crosses the x-axis at the point (4, 0). The gradient at the point is 1. The value of c is ?!?!?!?

please help me!

Answer 1

b=1
y=0
x=4
plug the values in the equation and get c

Answer 2

to find the point where a graph crosses 'x' axis, we put y=0
so, put y=0 in y=bx^2-cx and what you get ?

Answer 3

and since it crosses at 4,0, you put x=4.
you'll get one equation in b,c.

Answer 4

i will try thak you :)

Answer 5

ok, and to get other equation, differentiate the equation y=bx^2-cx
and put y'=gradient =1, and x=4.
Now you have 2 equations , 2 unknowns, easy to solve :)

Answer 6

but am i able to get variable free answer in this quetion ?

Answer 7

means ? show how much you could do ?

Answer 8

ahh dont worry! i got it lol

Answer 9

okay :)

Answer 10

thank you so much for your help!

Answer 11

welcome ^_^