Answer:
The other x-intercept is the point (1,0)
Step-by-step explanation:
we have
[tex]y=x^2-3x+2[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]y=x^2-3x+2[/tex]
equate to zero
[tex]x^2-3x+2=0[/tex]
so
[tex]a=1\\b=-3\\c=2[/tex]
substitute in the formula
[tex]x=\frac{-(-3)\pm\sqrt{-3^{2}-4(1)(2)}} {2(1)}[/tex]
[tex]x=\frac{3\pm\sqrt{1}} {2}[/tex]
[tex]x=\frac{3\pm1} {2}[/tex]
so
[tex]x=\frac{3+1} {2}=2[/tex]
[tex]x=\frac{3-1} {2}=1[/tex]
therefore
The other x-intercept is the point (1,0)
