Find the x-intercepts of the graph of the equation.Please help thanks(:

y=x^2-3x+2

y=x^2-3x+2 factor the quadratic
x^2-3x+2
(x-2)(x-1)
set these two to zero
(x-2)=0, (x-1)=0
solve for xs
1st gives x=2,
2nd gives x=1,
so these are the x-intercepts, x=2, x=1

you can factor this
(x-2)(x-1)=0 the numbers that make the equation =0, are your x-intercepts

x= 2, and x=1

---

y = (x-1)(x-2)

0 = (x-1)(x-2)

x = 1 OR x = 2

The x-intercepts are 1 and 2.

x^2 - 3*x + 2 = 0

(x - 2)*(x - 1) = 0

x1 = 2, x1 - int = (2, 0)

x2 = 1, x2 - int = (1, 0)

factorising y =(x-2)(x-1) now, x-intercepts are points where y=0
ie (x-2)(x-1) =0 ie x=2 and x=1

x^2 - 3x + 2 = 0
(x-2)(x-1) = 0
x = 2, 1
(2,0) and (1,0)