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)
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
2 answers
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)
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)
0 answers