Question

determine the maximum or the minimum value for each

a)y=2x^2+12

b)y=-3x^2-12x+15

c)3x(x-2)+5

Answer 1

If you're using calculus, just set y'=0 to find the max or min.

If you're using algebra, rearrange terms so can express the eaquation as

y = a(x-h)^2 + k

Then, the max or min is at (h,k)

For the first,

y = 2x^2 + 12

You can see that since x^2 is always at least zero, y has a minimum value when x=0. So, the minimum is at (0,12)


For the second,

y = 3x^2 - 12x + 15
= 3(x^2 - 4x + 5)
= 3(x^2 - 4x + 4) + 3
= 3(x-2)^2 + 3

so the minimum is at (2,3)


For the third,

y = 3x^2 - 6x + 5
= 3(x^2 - 2x) + 5
= 3(x^2 - 2x + 1) - 3 + 5
= 3(x-1)^2 + 2

so the minimum is at (1,2)

Answer 2

Oops, the second is a minus

y = -3x^2 - 12x + 15
= -3(x^2 + 4x - 5)
= -3(x^2 + 4x + 4 - 9)
= -3(x+2)^2 + 27

so the max is at (-2,27)