The graph of y=(px-1)/(2q+1) crosses the x-axis at (4,0) and the y-axis at (0,-5). How do find the values of p and q? So far, I've only substituted the x and y values in the equation...
The graph of y=(px-1)/(2q+1) crosses the x-axis at (4,0) and the y-axis at (0,-5). How do find the values of p and q? So far, I've only substituted the x and y values in the equation...
Two unknowns and two equations, simultaneous equations
If you substitute (4,0), you should notice something that in essence makes the q 'disappear' allowing you to solve for p and then q... (I think; I am very tired atm from AEA papers )
The graph of y=(px-1)/(2q+1) crosses the x-axis at (4,0) and the y-axis at (0,-5). How do find the values of p and q? So far, I've only substituted the x and y values in the equation...
thanks! could you show me how he wrote equation y=(px-1)/(2q+1) into mx + c
Lol sorry I messed up with the gradient. It should've been 0 - (-5) resulting in a positive 5/4 gradient and subsequently, a positive value for p. As for the mx + c part, you can turn any equation of a line into that form. Just make the equation into some kind of form where 'x' has a coefficient and 'y' is the subject of the equation. WHATEVER is being multiplied with 'x' will be the gradient 'm' and the rest will be 'c.'
Lol sorry I messed up with the gradient. It should've been 0 - (-5) resulting in a positive 5/4 gradient and subsequently, a positive value for p. As for the mx + c part, you can turn any equation of a line into that form. Just make the equation into some kind of form where 'x' has a coefficient and 'y' is the subject of the equation. WHATEVER is being multiplied with 'x' will be the gradient 'm' and the rest will be 'c.'
don't worry haha.. everyone makes mistakes....and thanks for the explanation! I understand it better now!