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
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)
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)
To determine the maximum or minimum value of a quadratic function, we can use the concept of vertex. The vertex represents either the maximum or minimum point on the graph of a quadratic function, depending on whether the parabola opens upward or downward.
a) For the function y = 2x^2 + 12, we can determine the minimum value. This is because the coefficient of x^2 is positive, indicating an upward-opening parabola. To find the minimum value, we need to find the vertex.
The vertex of a quadratic function in the form y = ax^2 + bx + c can be found using the formula:
x = -b / (2a)
For our function y = 2x^2 + 12, a = 2, and b = 0. Plugging these values into the formula, we get:
x = -0 / (2 * 2) = 0
To find the corresponding y-coordinate of the vertex, substitute the value of x back into the original equation:
y = 2(0)^2 + 12 = 0 + 12 = 12
Therefore, the minimum value of y for the function y = 2x^2 + 12 is 12. This occurs at the vertex (0, 12).
b) For the function y = -3x^2 - 12x + 15, we can determine the maximum value. This is because the coefficient of x^2 is negative, indicating a downward-opening parabola.
To find the maximum value, let's find the vertex using the formula x = -b / (2a):
a = -3 and b = -12, so we have:
x = -(-12) / (2 * -3) = 12 / (-6) = -2
Substituting -2 back into the original equation, we find the y-coordinate of the vertex:
y = -3(-2)^2 - 12(-2) + 15 = -3(4) + 24 + 15 = -12 + 24 + 15 = 27
Therefore, the maximum value of y for the function y = -3x^2 - 12x + 15 is 27. This occurs at the vertex (-2, 27).
c) For the function y = 3x(x - 2) + 5, there is no maximum or minimum value. This is because the coefficient of x^2 is zero, indicating a straight line, not a quadratic function.