Transform absolute-value graphs and find intersections

Build on the existing function lessons

Existing line and quadratic lessons supply equations, transformations, graph features and solving methods. Here the new operation is absolute value: it measures distance, changes a graph by folding, and requires branch conditions when solving intersections

Absolute value measures distance from zero

For real x, |x|=x when x≥0 and |x|=−x when x<0. Thus |x| is never negative. The graph consists of two rays meeting at(0,0), with slopes−1 and1. Its domain is R and range is[0,∞)

Evaluate |−7|−|3|

Separate the inside and outside transformations

For y=a|b(x−h)|+k with a,b nonzero, use |bu|=|b||u|. The effective multiplier is c=a|b|. The vertex is(h,k), left slope−c and right slope c. The range is y≥k if a>0 and y≤k if a<0. A negative b has no effect on this absolute-value graph

For y=−2|x−1|+3, which statement is correct?

Give the left branch slope of y=−3|−2(x+1)|+4.

A horizontal line can meet zero, one or two times

To solve a|x−h|+k=L, divide to get |x−h|=(L−k)/a. A negative right side gives no solution; zero gives the vertex once; a positive value d gives h−d and h+d. This remains true when a is negative

Worked level intersection

For−2|x+1|+3=−5, isolate |x+1|=4. Thus x=−5 or3. For the same graph at level3 there is one solution x=−1; at level4 there is none. The graph never rises above its vertex

How many distinct solutions does−2|x+1|+3=3 have?

An output fold changes negative y-values

For g(x)=|f(x)|, keep every part of f on or above the x-axis and reflect each negative part in that axis. Zeros stay fixed. The result is nonnegative, but it need not be V-shaped or one-to-one

What happens to y=x²−4 between−2 and2 under the output fold?

Evaluate |x²−4| at x=1.

An input fold copies the right half

For g(x)=f(|x|), keep f for x≥0 and reflect that half in the y-axis. The original left half is discarded. The resulting graph is even on its resulting domain, but its outputs can be negative. For a restricted original domain D, the new domain is the set of x with |x|∈D.

For f(x)=(x−2)²−1, which property of f(|x|) is shown?

Evaluate (|−2|−2)²−1.

Check nonnegative inner targets

To solve (|x|−H)²=r², first obtain |x|=H+r or H−r. Reject negative targets, count each positive target twice, and count a zero target once. For H=2,r=3 only |x|=5 remains. For H=r=2 the zeros are−4,0,4.

Intersect a V with a line branch by branch

Set u=x−h in m|x−h|+k=s(x−h)+k+q. On u≤0, solve(−m−s)u=q. On u≥0, solve(m−s)u=q. Retain only solutions in the relevant half-line. If both give u=0, it is one intersection

Worked two-branch intersection

For2|x−1|=x+3, the left branch gives−2(x−1)=x+3, so x=−1/3, valid since x≤1. The right gives2(x−1)=x+3, so x=5, valid since x≥1. Their y-values are8/3 and8.

Why are both displayed intersection candidates valid?

Give the larger solution of2|x−1|=x+3.

Parallel branches need a separate check

If a line has the same slope as a branch, its branch equation can become0=q. Then there are no solutions on that branch when q≠0, but the whole half-line when q=0. For |x|=x the solutions are all x≥0; there are infinitely many

What is the complete solution set of |x|=−x?

Two distance graphs can meet away from the midpoint

For |x|=2|x−3|, use x=2(x−3) or x=−2(x−3). These give x=6 and x=2. At6 the distances are6 and3; at2 they are2 and1. When the distance ratio is1, only the midpoint of two distinct fixed points remains

Give the smaller solution of |x|=2|x−3|

A folded quadratic can meet a level four times

For |x²−9|=5, solve x²=14 or x²=4, giving±√14 and±2. At level0, the two equations coincide and give only±3. At level9, they give x²=18 or0, hence three distinct roots. Above9 there are two; below0 there are none

How many distinct roots does |x²−9|=9 have?

A closed interval can hide the vertex

For y=−2|x−1|+3 on−1≤x≤2, endpoint values are−1 and1. The vertex x=1 is inside and has value3, so the range is[−1,3]. On2≤x≤4 the vertex is outside; endpoint values1 and−3 give range[−3,1]

Give the maximum of y=−2|x−1|+3 on−1≤x≤2.

Construct and justify a complete sketch, ungraded

Sketch y=−2|−3(x−1)|+4. Label the vertex, both intercepts, branch slopes, symmetry axis, domain and range. Explain why the sign of−3 does not reflect this graph. The complete sketch and explanation are ungraded

Model: reduce to a piecewise linear graph

The effective multiplier is−6, so the vertex is(1,4), slopes are6 and−6, and the symmetry axis isx=1. The y-intercept is−2. Setting the output to zero gives x-intercepts of one-third and five-thirds. The domain is R and the range is(−∞,4]. Draw two continuing rays through these data. Since |−3u|=3|u|, the negative inside sign disappears

Prove the branch rule, ungraded

For m>0, classify the solutions of m|u|=mu+q for q<0,q=0,q>0, including any infinite solution set. Prove that no candidate was omitted. This written argument is ungraded

Model: check both half-lines

For u≥0, the equation becomes0=q: it gives the whole nonnegative half-line only when q=0. For u<0, it becomes−2mu=q, so u=−q/(2m), valid exactly when q>0. Thus q<0 gives none, q=0 gives all u≥0, and q>0 gives exactly one negative solution. The two half-lines exhaust R and do not duplicate the origin

Compare folds and intersections with technology, ungraded

Plot f(x)=(x−2)²−1, |f(x)| and f(|x|). Explain their differing zeros, symmetry and negative outputs. Then compare |x²−9| with levels0,5,9 and10; verify the counts algebraically. This technology comparison is ungraded

Model: explain the operation before trusting the plot

The output fold keeps zeros1,3 and makes all outputs nonnegative; it is symmetric aboutx=2. The input fold has zeros−3,−1,1,3 and keeps negative minima; it is symmetric aboutx=0. For the folded quadratic, the four requested levels give2,4,3,2 intersections. A narrow window can hide outer roots, and numerical sampling can miss a tangent or repeated root, so use the equations to check the complete set