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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=4^2-4x-8
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You mean y=4x^2-4x-8point!? (2,0) means that x=2 when y=0, so you need to know the second salution, when y=zero. So find it with \[x_{1,2}=(-b \pm \sqrt{D})/2a = (4 \pm 12)/8 = 2; -1\]
the second point is (-1, 0)
x1 +x2 = -b/a....where c and a are coefficients this gives us x2 = 4/4 - 2
id21 is correct
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