Alan

In calculus the slope of a curve is obtained by a process called differentiation.  If you haven't done any calculus then clearly you can't obtain the maximum this way!  What way have you been taught to obtain the maximum?

Differentiate the expression with respect to x.  Set the result to zero and find the value of x which makes it zero.  Plug this value back in to the first expression to find the maximum value of the expression.

If the parabola has the x value of the vertex at 5, and a point on the x axis at x = 2  (which is 5-3), then it must also have a point on the x axis at (5+3,0) or at (8,0).

So you have the following two equations;

0 = a*2² + b*2 +c

0 = a*8² + b*8 + c

The slope at the vertex is zero, and the slope is given by dy/dx = 2a*x + b, so x = -b/2a.  We know this x is 5 when the slope is zero, so:

5 = -b/2a

You now have three equations to find the three unknowns, a, b and c.  

I'll leave the rest to you.

Hmm!

How are we to interpret 

as a cross-product, as element by element multiplication or what?

I get 598 as the largest.

Like so:

I'll let you take it from here.

FInd all k for which x^2 + 5kx + k^2 + 5 is divisible by x + 2, but not by x + 3.

As follows:

Correction:  The final area should just be 9, of course, not 9/2.

An equilateral triangle with side length s is positioned with one vertex on the center of a circle with radius r. Find the quotient s/r that causes the area inside the triangle but outside the circle to be the same as the area of the circle.