Okay, let’s look at it this way: when does a line pass through the origin? The line represents possible values of x and y that satisfy the equation.
So, when a line passes through the origin, it passes through the coordinates (0,0). x = 0, y = 0. So, let’s model this with the equation ax + by + c = 0. Sub in x = 0 and y = 0 to the equation and we find that 0 + 0 + c = 0. Clearly, c = 0.
So, with this simple explanation, I hope you understand when a line does pass through the origin.
Now, let’s look at when a line doesn’t pass through the origin. This is when c is not equal to 0. Hence, when x = 0, y cannot equal 0; c + by = 0, and we know that c is not 0. If y is 0, then we get c = 0… where c is not 0. Ehh. Thus, you can see that a line does not pass through the origin when c is not equal to 0 by ehat is hope is a simple explanation. You don’t need to know how to prove it, I presume, but that’s not too hard either.
Oops, I realised that I just assumed you were talking about linear graphs. For quadratic graphs, the reasoning is similar. For graphs of the form y = ax^2, the minimum/maximum point of the graph will be the origin. For graphs of the form y = ax^2 + bx, it will pass through the origin but the line of symmetry will be different. For graphs of the form y = ax^2 + bx + c (you know, where c is not zero) , the graph will not pass through the origin because the maximum/ minimum point is actually raised or lowered by c units
When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
