Question

One x-intercept for a parabola is at the point

(2,0). Use the quadratic formula to find the
other x-intercept for the parabola defined by
this equation:
y=x² – 3x + 2

Answer 1

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

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√(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)