SOLUTION: I have to find where the parabola crosses the x-axis and the y-axis for this equation: Y=x^2-3x+2. Could someone please help?

Question 875463: I have to find where the parabola crosses the x-axis and the y-axis for this equation: Y=x^2-3x+2. Could someone please help?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
y = x^2 - 3x + 2
to find the y-axis crossing, make the value of x = to 0 and solve for y
you get:
y = 0^2 - 3*0 + 2 which is equal to 2.
the graph crosses the y-axis as (0,2).
to find the x-axis crossing, you have to set y = 0 and solve the equation.
you get:
0 = x^2 - 3x + 2
factor this equation to get:
(x-2) * (x-1) = 0
set each of these factors equal to 0 to get:
x - 2 = 0
x - 1 = 0
solve for x in each of these equations to get:
x = 2
x = 1
the equation crosses the x-axis at (2,0) and (1,0)
a graph of your equation is shown below:

you can see that it crosses the y-axis at y = 2 and the x-axis as x = 1 and x = 2
the coordinate points are:
(0,2), (1,0), (2,0)

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