You have the following function:
f(x) = x² - x
- In order to determine the vertex of the function, consider that for the general form of a quadratic function:
f(x) = ax² + bx + c
the value of x at the vertex is:
x = -b/2a
for the given function you have a = 1, b = -1, c =0:
x = -(-1)/2(1) = 1/2 = 0.5
next, replace the previous value of x into the function:
f(1/2) = (1/2)² - (1/2) = 1/4 - 1/2 = -1/4 = -0.25
Hence, the vertex is (0.5 , -0.25)
- The axis of symmetry is the value of x atthe vertex:
x = 0.5
- The x-intercepts are the zeros of the function:
x² - x = 0 by factorizing
x(x - 1) = 0
Then, the zeros are:
x =0
x = 1
Hence, the x-intercepts are x=0 and x=1
- Due to the coefficient a is positive and the term ax² is the dominant term, this curve has a minimum. This minimum is the vertex (0.5 , -0.25)
- The minimum value of the function is -0.25
- The y-intercept is the value of y when x=0:
f(0) = 0² - 0 = 0