Answer:
1 and 2
Step-by-step explanation:
Zeros of the function is where the function intersect with x-axis (or in other way the zeros of function is x value when y = 0 )
y=x^2-3x+2
(we substitute y with zero)
0 = x^2-3x+2
or in other way
x^2-3x+2 = 0
Then we factor out the function by using the grouping method
(When we should give two numbers that there sum is equal -3 and if we multiply them we get 2)
and these two numbers are -2 and -1
x^2-3x+2 = 0
*Substitute the middle term with -x-2x
x^2-x-2x+2 = 0
*factor out x , then factor out -2
x(x-1)-2(x-1) = 0
*factor out x-2
(x-2)(x-1)=0
Either
x-2 = 0
So
x = 2
Or
x-1 = 0
So
x = 1
in conclusion the zeros of this function is 1,2