x-intercepts of parabola: y = 0
We have y = 2x² - 8x + 8
[tex]y=0\to2x^2-8x+6=0\ \ \ \ |:2\\\\x^2-4x+3=0\\\\x^2-3x-x+3=0\\\\x(x-3)-1(x-3)=0\\\\(x-3)(x-1)=0\iff x-3=0\ \vee\ x-1=0\\\\\boxed{x=3\ \vee\ x=1}[/tex]
Answer: x-intercepts: x = 1 and x = 3.
x -intercepts are (1 , 0) and (3 , 0)
to find the x-intercepts let y = 0, thus
2x² - 8x + 6 = 0 → ( take out a common factor of 2 )
2(x² - 4x + 3 ) = 0 → ( factor the quadratic )
the factors of + 3 which sum to - 4 are - 1 and - 3, hence
2(x - 1)(x - 3) = 0
equate each factor to zero and solve for x
x - 1 = 0 → x = 1 ⇒ (1 , 0)
x - 3 = 0 → x = 3 ⇒ (3 , 0) are the x- intercepts
