Question

One x-intercept for a parabola is at the point

Answer 1

Answer:

The other x-intercept is the point (1,0)

Step-by-step explanation:

we have

$y=x^2-3x+2$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$y=x^2-3x+2$

equate to zero

$x^2-3x+2=0$

so

$a=1\\b=-3\\c=2$

substitute in the formula

$x=\frac{-(-3)\pm\sqrt{-3^{2}-4(1)(2)}} {2(1)}$

$x=\frac{3\pm\sqrt{1}} {2}$

$x=\frac{3\pm1} {2}$

so

$x=\frac{3+1} {2}=2$

$x=\frac{3-1} {2}=1$

therefore

The other x-intercept is the point (1,0)