y = 2x^2 + x -1
If you know calculus you can find the minimum or maximum when the slope of the equation is zero. This is the first derivative of y with respect to x
dy/dx = 4x +1 = 0
4x = -1
x = -1/4
y = 2 (-1/4)^2 -1/4 -1 = 1/8-1/4-1 = 1/8 - 2/8 - 8/8 = -9/8 minimum = -1.125
The vertex has a minimum value of (-4, -11)
No. By definition, a function has a single unique value for every value passed into it. The equation given here describes a circle, which can not be rearranged to meet this condition.
The function that is given has a constant value and therefore, its slope is 0.
It has an absolute minimum at the point (2,3). It has no maximum but the ends of the graph both approach infinity.
The function f(x) = 3x + 2 can take any real value, so the question does not seem to make any sense.
The answer will depend on the ranges for x and y. If the ranges are not restricted, then C can have any value.
The vertex has a minimum value of (-4, -11)
You can tell if an equation is a function if for any x value that you put into the function, you get only one y value. The equation you asked about is the equation of a line. It is a function.
5 7
No. By definition, a function has a single unique value for every value passed into it. The equation given here describes a circle, which can not be rearranged to meet this condition.
The function that is given has a constant value and therefore, its slope is 0.
y=2x2+5. Plug in 0 for x, and you get y=2(0)^2 +5 =0+5=5. 5 is the minimum value for y.
If a + b = 6, what is the value of 3a + 3b?
The function f(x) = 3x + 2 can take any real value, so the question does not seem to make any sense.
It has an absolute minimum at the point (2,3). It has no maximum but the ends of the graph both approach infinity.
It is a quadratic function which represents a parabola.
The minimum value of the parabola is at the point (-1/3, -4/3)