o Absolute Value Rule
-> Absolute value measures distance from zero, never negative.
-> |x| = x if x >= 0.
-> |x| = -x if x < 0.
-> For example, |-5| = -(-5) = 5 and |5| = 5.
-> The parent graph is a V with vertex (0, 0).
-> Its domain is all real numbers; its range is y >= 0.
-> Opposite inputs have equal absolute values.
-> |x - c| is the distance between the input and the point c.
-> Factoring: |2x - 6| = 2|x - 3| leaves the zeros unchanged.
o General Graph Form
-> Write y = a|x - h| + k, with a != 0.
-> The vertex is (h, k); the symmetry line is x = h.
-> The domain is all real numbers.
-> If a > 0, the V opens upward and the range is y >= k.
-> If a < 0, the V opens downward and the range is y <= k.
-> The left branch slope is -a; the right branch slope is a.
-> Larger |a| makes a steeper V; smaller |a| makes it flatter.
-> The V meets the x-axis exactly when a*k <= 0.
-> Points the same distance from h give the same output y.
o Upward Worked Example
-> For y = |x - 1| + 3, a = 1, h = 1, and k = 3.
-> The vertex is (1, 3); the symmetry line is x = 1.
-> The minimum output is 3; the range is y >= 3.
-> At x = 0 or x = 2, y = 4.
-> At x = -1 or x = 3, y = 5.
-> For x < 1, y = -(x - 1) + 3 = -x + 4.
-> For x >= 1, y = x - 1 + 3 = x + 2.
-> x-intercepts: |x - 1| = -3 has none, so no crossings.
o Downward Worked Example
-> For y = -(2/3)|x + 3| + 1, h = -3, not 3.
-> The vertex is (-3, 1); the symmetry line is x = -3.
-> Since a = -2/3, the maximum output is 1.
-> The domain is all real numbers; the range is y <= 1.
-> The left slope is 2/3; the right slope is -2/3.
-> At x = -6 or x = 0, y = -(2/3)(3) + 1 = -1.
-> Plot these symmetric points and draw the two rays.
-> At x = -3 the absolute term vanishes, leaving y = 1.
-> The y-intercept is (0, -1), one of the symmetric points.
o Equation Strategy
-> Isolate the absolute value before splitting into cases.
-> For |u| = q with q < 0, there are no real solutions.
-> For q = 0, solve u = 0; for nonconstant linear u, one root.
-> For q > 0, solve u = q or u = -q.
-> If u is linear with nonzero slope, q > 0 gives two roots.
-> For nonlinear u, the number of roots can differ.
-> Check roots in the original equation for extraneous roots.
-> Graph view: |u| = q cuts the curve y = |u| at height q.
-> After splitting, keep each branch with its case condition.
o Solve and Check
-> Solve 2|x - 1| + 3 = 11: first get |x - 1| = 4.
-> x - 1 = 4 or x - 1 = -4 gives x = 5 or x = -3.
-> Check both: 2(4) + 3 = 11, so both are solutions.
-> |x + 2| = -3 has no solutions: distance cannot be negative.
-> |x + 2| = 0 gives x = -2, exactly one solution.
-> Squaring |x + 2| = -3 would falsely suggest x = 1 or -5.
-> Both fail the original equation; reject those extra roots.
-> |x - 2| = 5 gives x = 7 and x = -3, both checked.