For the function f(x) = x^2 - 3x + 1, at the x - intercept f(x) = 0

0 = x^2 - 3x + 1

x1 = 3/2 + [sqrt (9 - 4)] / 2

=> x1 = 3/2 + (sqrt 5)/2

x2 = 3/2 - (sqrt 5)/2

The x-intercepts are (3/2 + (sqrt 5)/2 , 0) and (3/2 - (sqrt 5)/2, 0)

At the y-intercepts x = 0

=> f(x) = 0 - 0 + 1 = 1

The y intercept is (0,1)

**The x intercepts are (3/2 + (sqrt 5)/2 , 0) and (3/2 - (sqrt 5)/2, 0) and the y intercept is (0, 1)**

To determine x intercept, we'll put y = f(x) = 0.

But f(x) = x^2-3x+1 => x^2-3x+1 = 0

We'll apply quadratic formula:

x1 = [3+sqrt(9 - 4)]/2

x1 = (3+sqrt5)/2

x2 = (3-sqrt5)/2

The x intercepts are: ((3+sqrt5)/2 , 0) and ((3-sqrt5)/2) , 0).

To determine y intercepts, we'll cancel the x coordinate.

f(0) = 0^2 - 3*0 + 1

f(0) = 1

The y intercept is (0 , 1).

**The x intercepts are ((3+sqrt5)/2 , 0) and ((3-sqrt5)/2) , 0) and the y intercept is (0 , 1).**