Learning goals: Build linear rules; interpret piecewise definitions; describe transformations using input-output reasoning.
Linear structure
A linear function has the form \(f(x)=mx+b\). Its slope \(m\) is the change in output per unit change in input, and \(b=f(0)\). Through \((2,7)\) and \((5,16)\), the slope is \((16-7)/(5-2)=3\). Point-slope form gives \(y-7=3(x-2)\), or \(y=3x+1\). Parallel nonvertical lines have equal slopes. Perpendicular nonvertical lines have slopes whose product is negative one. Horizontal and vertical lines form the separate perpendicular pair; a vertical line is not a function of x.
The absolute-value parent
The rule \(|x|\) equals \(x\) for nonnegative inputs and \(-x\) for negative inputs. Its V-shaped graph records distance from zero. Consequently, \(g(x)=2|x-3|-4\) has vertex \((3,-4)\), opens upward, and has arms with slopes negative two and positive two. Its range is \([-4,\infty)\). To find its zeros, solve \(2|x-3|-4=0\), so \(|x-3|=2\) and \(x=1,5\).
Transformations act on inputs and outputs
In \[g(x)=a\,f(b(x-h))+k,\qquad b\ne0,\] the graph is shifted horizontally by \(h\) and vertically by \(k\). The magnitude of \(a\) changes vertical distances; a negative \(a\) reflects across the x-axis. The magnitude of \(b\) scales horizontal distances by \(1/|b|\); negative \(b\) reverses them. The most reliable rule tracks a point: \[(u,v)\text{ on }f\quad\longmapsto\quad(h+u/b,\;av+k)\text{ on }g.\] This avoids memorizing “inside does the opposite” without understanding why.
For \(g(x)=-2f(x-3)+1\), a point \((4,5)\) becomes \((7,-9)\). The input must be seven so that \(x-3=4\); the output then becomes \(-2(5)+1=-9\). For \(f(2x-6)=f(2(x-3))\), the horizontal shift is three, not six. Factor the entire inside expression first.
Piecewise rules
A piecewise function chooses a formula according to the input. Suppose shipping costs \[S(w)=\begin{cases}5,&0\lt w\le2,\\5+2(w-2),&2\lt w\le5.\end{cases}\] The input is package weight in kilograms, and the output is dollars. At \(w=2\), use the first rule, obtaining five dollars. At \(w=3\), use the second, obtaining seven dollars. Do not evaluate every branch and combine the results. Endpoint conditions determine which graph points are filled or open.
Compare representations
Draw the parent graph, mark key points, transform those points, and then connect them with the correct shape. Check the vertex or intercepts algebraically. A calculator window can hide an important feature, so choose a range that shows the landmarks you have already predicted. Explain which feature changed and which remained unchanged.
Practice: attempt before reading the solutions
Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.
- Find the line through \((1,2)\) and \((4,11)\).
- Find a slope perpendicular to \(y=\frac23x+7\).
- Give the vertex and range of \(-|x+2|+5\).
- Solve \(3|x-1|=12\).
- If \((2,7)\) lies on \(f\), find its transformed point for \(g(x)=3f(x+4)-2\).
- Describe the horizontal changes in \(f(2x-8)\).
- Let \(p(x)=x+1\) for \(x\lt0\), and \(p(x)=x^2\) for \(x\ge0\). Find \(p(-2),p(0),p(3)\).
- Does \(y=5\) define a function of x? Does \(x=5\)?
Worked solutions and reasoning
Compare these steps with your own work. The solution numbers match the practice questions above.
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The slope is \((11-2)/(4-1)=3\). Using \((1,2)\), \(2=3+b\), so \(y=3x-1\).
The change in y is nine and the change in x is three. After finding slope three, use \(y=3x+b\) and one given point to find b. Check the other point: \(3(4)-1=11\).
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The negative reciprocal is \(-3/2\). Its product with \(2/3\) is \(-1\).
For two nonvertical perpendicular lines, their slopes multiply to \(-1\). Solving \((2/3)m=-1\) gives \(m=-3/2\). The intercept seven affects the line’s position, but it does not affect its slope.
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The vertex is \((-2,5)\). The negative sign makes five the maximum, so the range is \((-\infty,5]\).
The absolute value is smallest, namely zero, at \(x=-2\). Negating it makes every other contribution nonpositive, so the output cannot exceed five. Outputs decrease without bound as the distance from \(-2\) increases.
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Divide by three to obtain \(|x-1|=4\). Then \(x-1=4\) or \(-4\), so \(x=5\) or \(-3\).
The equation \(|x-1|=4\) asks for inputs at distance four from one. These are \(1+4=5\) and \(1-4=-3\). Each gives \(3\cdot4=12\) in the original equation.
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Input \(-2\) makes \(x+4=2\); output \(3(7)-2=19\). The point is \((-2,19)\).
The original point means \(f(2)=7\). To use this known value in g, solve \(x+4=2\) for the new input. Then apply the outside operations to the original output. Inside changes affect x; outside changes affect y.
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Factor as \(f(2(x-4))\). Horizontal distances are scaled by \(1/2\), and the graph is shifted right four units.
An original point \((u,f(u))\) appears at the new input satisfying \(2x-8=u\), so \(x=u/2+4\). This point rule makes the order clear: halve the original x-coordinate, then add four; the output remains unchanged.
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Choose branches by the input: \(p(-2)=-1\), \(p(0)=0\), and \(p(3)=9\). Zero belongs only to the second branch.
The branch condition is tested before evaluating the formula. Negative two uses \(x+1\), whereas zero and three use \(x^2\). Using the first formula at zero would give one, but that branch explicitly excludes zero.
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The horizontal line \(y=5\) assigns one output to each input and is a function. The vertical line \(x=5\) allows many outputs for one input and fails the vertical line test.
For \(y=5\), even very different x-values still receive exactly one output each. For \(x=5\), both \((5,0)\) and \((5,1)\) lie on the relation, giving the same input different outputs. That violates the function definition.
Reflection
Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.