|
No, both terms include X so there can be no intercept other than the origin for a truly linear equation, however this isn't linear.
normally you have y = mx + c, which gives the intercept as c, in this case we have two X terms, when we evaluate this as X -> 0 then we get 0a + b/+-0 (a very small number), as such at the origin the actual value is undefined, similar to a tan curve at various points.
For large values of X aX dominates the equation and we have a function which says Y^2 = aX or Y = (aX)^1/2, and again for negative X, at the origin the b/X dominates and we have the graph rapidly shooting off to +ve/-ve infinity for +0/-0.
|