Let 1f(x)=x^2-2bx b is a constant.
Determine the following in terms of b. (this means there can be b's in your answers)
Zeros (x-intercepts)
y-intercept
Axis of Symmetry
Extreme Value
To find the zeros (x-intercepts), set f(x) to 0 and solve for x:
0 = x^2 - 2bx
x^2 = 2bx
x = ±√(2bx)
So the zeros are x = ±√(2bx).
To find the y-intercept, set x to 0:
f(0) = 0^2 - 2b(0) = 0
So the y-intercept is (0,0).
To find the axis of symmetry, recall that the axis of symmetry for a quadratic function in the form of f(x) = ax^2 + bx + c is given by the formula:
x = -b / 2a
In this case, a = 1 and b = -2b:
x = -(-2b) / 2(1) = (2b) / 2 = b
So the axis of symmetry is x = b.
The extreme value occurs at the vertex of the parabola. Since the coefficient of the x^2 term is positive, the parabola opens upwards, so the extreme value is a minimum value. The vertex is located at the axis of symmetry, which is x = b. To find the y-coordinate of the vertex, plug the x-coordinate into the function:
f(b) = b^2 - 2b*b = b^2 - 2b^2 = -b^2
So the extreme value is the minimum value, and it occurs at the point (b, -b^2).