find the maximum or minimum if the quadratic function?

To find maximum/minimum points you need to find the first and second derivatives of f(x)

dy/dx=-6x+12, and d2y/dx2=-6, this means that there is only an absolute maximum when dy/dx=0

dy/dx=0 when 6x=12, x=2

f(2)=-3(4)+12(2)-2=10

So it has an absolute maximum at (2,10) and it decreases without bound as x->±infinity...

...

dy/dx=-2x-2, d2y/dx2=-2, so like the last problem this function only has an absolute maximum when dy/dx=0 because the acceleration, d2y/dx2, is always negative...

dy/dx=0 when -2x-2=0, -2x=2, x=-1

f(-1)=-(-1)^2-2(-1)+8=9

So this function has only an absolute maximum at (-1,9) and decreases without bound as x->±infinity...

1.)

f(x) = -3x^2 + 12x - 2

= -3(x^2 - 4x + 2/3)

= -3((x - 2)^2 - 10/3)

= -3(x - 2)^2 + 10

So the maximum is 10

2.)

y = -(x^2 + 2x - 8)

y = -((x + 1)^2 - 9)

y = -(x + 1)^2 + 9

So the maximum is 9

If you learned derivatives in your course:

- find the first derivative;

- equal it to zero;

- solve it for x

This is your max or min

If you didn't study derivatives, find the vertex of parabola.