Find the maximum or minimum of the following quadratic function: y = x2 - x - 56.
A. -56
B. 56
C. -225/4
D. 225/4
d
1. y=x2 - 16
for maximum or minumum, dy/dx=0
hence 2x=0 or x=0
minimum value of function occurs when x=0 which is y=-16
2.using same principle as above,
dy/dx=10x+5
when dy/dx=0 => 10x+5=0 => x=-1/2
minimum value of y= -5/4 - 5/2 + 11
for 3, 4 and 5 use the same principle as 1 and 2
max: infinity (at x=inf)
min: x=1/2, y= you figure it. check sign.
Not sure if you know Calculus, as done by Me
so let's complete the square
y = x^2 - x - 56
= x^2 - x + 1/4 - 1/4 - 56
= (x- 1/2)^2 - 225/4
so we have a min of -225/4 , when x = 1/2
(min because the parabola opens upwards)
To find the maximum or minimum of a quadratic function, you need to recognize that it is in the form of "y = ax^2 + bx + c," where "a," "b," and "c" are coefficients.
In this case, the quadratic function is y = x^2 - x - 56. We can identify that "a" is 1, "b" is -1, and "c" is -56.
To find the maximum or minimum, we can use the formula x = -b / 2a, which gives you the x-coordinate of the vertex of the quadratic function.
In this case, the formula becomes x = -(-1) / (2 * 1) = 1/2.
Next, substitute this x-coordinate back into the original equation to find the y-coordinate of the vertex:
y = (1/2)^2 - (1/2) - 56
y = 1/4 - 1/2 - 56
y = -7/4
Thus, the coordinates of the vertex are (1/2, -7/4).
To determine if the vertex is a maximum or minimum, we need to look at the coefficient of "a." Since "a" is positive (1 in this case), the parabola opens upwards, and the vertex represents a minimum point.
Therefore, the minimum value of the quadratic function y = x^2 - x - 56 is -7/4, which corresponds to option D, 225/4.