Recent content by Gary Smart

Does the discriminant check work because we are trying to find where the combined function of ax^2 + bx + c = mx + c crosses the x axis?

If parameter a is positive and the parabolic function has a vertex the two points of intersection, the only way to find where it lies is by completing the square (using vertex form)?

Thank you very much for your help. I'm looking at different y intercepts now, to test for a lower bound that is not 0. I know the Y intercepts will be different => ax^2 + bx + c = mx + d That equation can not be simplified.. does this mean there is not a general case when the y intercept...

Yes, I understand that. This is because if it wasn't the x value would be negative and would not be in the first quadrant. Therefore, if the gradient of the line and the gradient of the circle subtract to give a negative number then a must be negative and vice versa?

The general case for finding the x coordinates of the intercepts is calculated by the following: Setting: ax^2 + bx + c equal to mx + c => ax^2 + bx + c = mx + c Solving for x gives: x = 0 and x = m − b / a This means that the points of interception between a parabola and a line will always be...

Hello Sammy, firstly thank you for helping me. I have been working with f1(x)=mx+c and f2(x)=ax2+bx+c and trying different values. The problem I'm finding is, if I find the other point of intersection Xr (which is dependant on the functions) then when I alter the values of the equations...

Momoko. I'm a little lost with that. Where do we get ax^2+(b-m)x from? Could you simplify it a little further for me please? I know that the parabola and the line have to intersect, so we set them equal to each other. The x values are the coordinates of intersection which then become the...

I've done some basic calculus. Ah, yes I definitely understand that but I don't understand "Would you have been happier if you were instead given a equation in m that was equal to 1? Because that is what the integral is! It's just sort of hiding the answer, waiting for you to solve it so it can...

The mental block that I'm finding difficult to get past is the idea of letting an integral = 1. It makes no sense to me.

I understand now, just had a little misunderstanding in the previous text. I I have found the line that intersects the parabola [y = (x-1)^2]. It is approximately: y = 0.702209x The points of intersection are 2.25967 and 0.442543. These values seem very far fetched though, are they supposed...

The level of maths I have studied is GCSE (high school), that was a few years ago now though. This challenge I came across individually and I'm really interesting in investigating it and solving it. So, I took the parabola: y = (x - 1)^2. I chose a line that intercepts it. The line was 2x. The...

Just to make sure I'm clear. That equation involves integrating the general form of a line and parabola simultaneously. The u and v values are the points of interception as we are trying to find the area between these bounds. How do I solve for x though with no values?

I've carried out some investigations involving altering values of the quadratic then change the gradient of the line to get an area of 1. I think all the combinations of the quadratic can fit into the investigations carried out in that table. I haven't really seen any patterns though...

Ahhh. I see. So there isn't a way of finding an area of 1 with these restrictions using algebra?

So, I need to apply the trapezoidal rule in general to general forms of both types of functions?