SOLUTION: in y=ax^2+bx+c i understand that b is a strong determinant of the x-intercepts of the curve, but can you explain that more to me? and please give examples if possible!

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: in y=ax^2+bx+c i understand that b is a strong determinant of the x-intercepts of the curve, but can you explain that more to me? and please give examples if possible!      Log On

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Question 70739: in y=ax^2+bx+c i understand that b is a strong determinant of the x-intercepts of the curve, but can you explain that more to me? and please give examples if possible!
Answer by josmiceli(6784) About Me  (Show Source):
You can put this solution on YOUR website!
The roots, or x-intercepts, are found by the formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
rewrite this as
x+=+%28-b+%2F+2a%29+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29+%2F+%282%2Aa%29+
The b%5E2+-+4ac under the square root sign
determines the nature of the x-intercepts (roots).
If b%5E2+=+4ac, the whole 2nd term is zero, and
there is 1 x-intercept at -b%2F%282a%29. All this
means is that the vertex of the parabola just kisses
the x-axis.
If b%5E2+%3E+4ac, There are 2 x-intercepts at these locations:
x+=+%28-b+%2F+2a%29+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29+%2F+%282%2Aa%29+
and
x+=+%28-b+%2F+2a%29+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29+%2F+%282%2Aa%29+
If b%5E2+%3C+4ac, There are a pair of imaginary roots
(involving sqrt%28-1%29)
All that means is that the parabola never touches the x-axis