We need the Quadratic formula for this y=(x-2)(x-6)
My daughter is working on this problem and we can't figure this out.
y = (x - 2)(x - 6)
y = x^2 - 2x - 6x + 12
y = x^2 - 8x + 12
An equation of this type defines a parabola. What we are looking for
is any points where the graph crosses the x-axis. If those points exist,
the y value is by definition zero. Therefore, we set the equation equal to 0.
x^2 - 8x + 12 = 0
This is in the form ax^2 + bx + c = 0. The a, b and c are the values
used in the quadratic formula.
-b ± √(b^2 - 4ac)
x = -----------------------
2a
-(-8) ± √((-8)^2 - 4(1)(12))
x = ----------------------------------
2(1)
8 ± √(64 - 48)
x = -------------------
2
8 ± √(16)
x = -------------
2
8 ± 4
x = -------
2
8 + 4 8 - 4
x = ------- and x = --------
2 2
12 4
x = ----- and x = ----
2 2
x = 6 and x = 2
These values mean that the graph crosses the
x-axis at (2, 0) and (6, 0)