Determine whether the quadratic function has a minimum or maximum value.Then find the coordinates of the minimum or maximum point.
f(x)=2x^2-4x
To determine whether the quadratic function has a minimum or maximum value, we need to analyze the coefficient of the x^2 term. In this case, the coefficient is positive (2), which means that the graph of the quadratic function opens upward.
Since the quadratic function opens upward, it has a minimum value. To find the coordinates of the minimum point, we need to use the vertex formula.
The formula for the x-coordinate of the vertex is given by:
x = -b / (2a)
In the function f(x) = 2x^2 - 4x, the coefficient of x^2 is a = 2, and the coefficient of x is b = -4. Plugging these values into the vertex formula, we get:
x = -(-4) / (2 * 2)
x = 4 / 4
x = 1
To find the y-coordinate of the vertex, we substitute the value of x into the function:
f(1) = 2(1)^2 - 4(1)
f(1) = 2 - 4
f(1) = -2
Therefore, the coordinates of the minimum point are (1, -2).
To determine whether the quadratic function has a minimum or maximum value, we can look at the coefficient of the x^2 term. In this case, the coefficient is positive (2), which means the parabola opens upwards and has a minimum value.
To find the coordinates of the minimum point, we can use the vertex formula. The vertex formula for a quadratic function in the form ax^2 + bx + c is given by:
x = -b / (2a)
In this case, a = 2 and b = -4. Plugging these values into the formula, we can find the x-coordinate of the vertex:
x = -(-4) / (2*2) = 4 / 4 = 1
To find the y-coordinate of the vertex, we substitute the x-coordinate back into the original function:
f(1) = 2(1)^2 - 4(1) = 2 - 4 = -2
Therefore, the coordinates of the minimum point are (1, -2).
y = 2 (x^2-2x)
dy/dx = 2 (2x-2) = extreme at x = 1
d^2y/dx^2 = 4 so that extreme is a minimum
at x = 1
f(1) = -2
if you know Calculus, then
f'(x) = 4x - 4
= 0 for a max/min
4(x-1) = 0
x = 1
then f(1) = 2-4 = -2
so (1,-2) is a min (since the parabola opens upwards)
if you don't know Calculus, use "completing the square" to find the vertex
f(x) = 2[x^2 - 2x + 1 - 1 ]
= 2[(x-1)^2 - 1]
= 2(x-1)^2 - 2
so the vertex is (1,-2) as above
or
you could use the formula:
the x of the vertex is -b/(2a)
= -(-4)/2(2)
= -1
etc