Graphs of Absolute Value Functions

# Graphs of Absolute Value Functions ## Objectifs et prérequis This chapter focuses on the graphical representation of absolute value (modulus) functions, a fundamental topic in Lower Sixth Mathematics within the Cameroonian Anglophone education system. By the end of this chapter, students will be able to: - Define the absolute value function and interpret its geometric meaning as distance from zero on the real number line. - Sketch graphs of basic absolute value functions using transformations (translations, reflections, stretches). - Solve equations and inequalities involving absolute value functions graphically and algebraically. - Analyse piecewise definitions of absolute value functions and construct their graphs accordingly. - Apply absolute value functions to model real-world situations involving distance, error, or deviation. **Prerequisites**: - Proficiency in solving linear equations and inequalities. - Understanding of function notation, domain, and range. - Familiarity with basic transformations of functions (shifts, reflections, scaling). - Ability to sketch graphs of linear and quadratic functions. - Knowledge of the Cartesian coordinate system and plotting points. --- ## I. Graphs of modulus or absolute value functions 1 ### Definition and Basic Properties The absolute value (or modulus) of a real number $x$, denoted $|x|$, is defined as: $$ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} $$ This definition ensures $|x| \geq 0$ for all real $x$, and $|x| = 0$ if and only if $x = 0$. > **Key Property 1**: $|x|$ represents the distance of $x$ from 0 on the number line. > **Key Property 2**: $|x| = \sqrt{x^2}$ for all real $x$. > **Key Property 3**: $|xy| = |x||y|$ and $\left|\frac{x}{y}\right| = \frac{|x|}{|y|}$ for $y \neq 0$. > **Key Property 4 (Triangle Inequality)**: $|x + y| \leq |x| + |y|$, with equality if and only if $xy \geq 0$. The graph of the basic absolute value function $f(x) = |x|$ is a V-shaped curve symmetric about the $y$-axis, with its vertex at the origin $(0,0)$. It consists of two linear pieces: - For $x \geq 0$, $f(x) = x$ (a line with slope 1). - For $x < 0$, $f(x) = -x$ (a line with slope $-1$). ### Transformations of $f(x) = |x|$ We extend our understanding by applying transformations to the parent function $f(x) = |x|$. The general form is: $$ g(x) = a|bx + c| + d $$ where: - $a$ controls vertical stretch/compression and reflection over the $x$-axis (if $a < 0$). - $b$ controls horizontal stretch/compression and reflection over the $y$-axis (if $b < 0$). - $c$ induces a horizontal shift: left by $\frac{c}{|b|}$ if $c > 0$, right if $c < 0$. - $d$ induces a vertical shift: up by $d$ if $d > 0$, down if $d < 0$. The vertex of $g(x) = a|bx + c| + d$ is located at $\left(-\frac{c}{b}, d\right)$, provided $b \neq 0$. The axis of symmetry is the vertical line $x = -\frac{c}{b}$. #### Example 1: Graph $g(x) = 2|x - 3| + 1$ **Step 1**: Identify transformations from $f(x) = |x|$. - $a = 2$: vertical stretch by factor 2. - $b = 1$: no horizontal stretch/compression or reflection. - $c = -3$: rewrite as $|1\cdot x + (-3)|$, so horizontal shift right by 3 units (since $-\frac{c}{b} = -\frac{-3}{1} = 3$). - $d = 1$: vertical shift up by 1 unit. **Step 2**: Find vertex: $\left(-\frac{c}{b}, d\right) = (3, 1)$. **Step 3**: Determine slopes of the arms. - For $x \geq 3$: $g(x) = 2(x - 3) + 1 = 2x - 6 + 1 = 2x - 5$, slope = 2. - For $x < 3$: $g(x) = 2(-(x - 3)) + 1 = -2x + 6 + 1 = -2x + 7$, slope = $-2$. **Step 4**: Plot key points. - Vertex: $(3, 1)$. - When $x = 4$: $g(4) = 2|4-3|+1 = 2(1)+1=3$ → $(4,3)$. - When $x = 2$: $g(2) = 2|2-3|+1 = 2(1)+1=3$ → $(2,3)$. - When $x = 0$: $g(0) = 2|0-3|+1 = 2(3)+1=7$ → $(0,7)$. - When $x = 6$: $g(6) = 2|6-3|+1 = 2(3)+1=7$ → $(6,7)$. **Step 5**: Sketch the V-shaped graph with vertex at $(3,1)$, opening upwards, arms with slopes $\pm 2$. > **To Remember**: The vertex of $a|x - h| + k$ is at $(h, k)$, and the slopes of the arms are $\pm a$ (if $b=1$). #### Example 2: Graph $h(x) = -|2x + 4| - 3$ **Step 1**: Rewrite in standard form: $h(x) = -|2(x + 2)| - 3$. - $a = -1$: reflection over $x$-axis. - $b = 2$: horizontal compression by factor $\frac{1}{2}$. - $c = 4$: equivalent to shift left by 2 units (since $-\frac{c}{b} = -\frac{4}{2} = -2$). - $d = -3$: vertical shift down by 3 units. **Step 2**: Vertex: $\left(-\frac{c}{b}, d\right) = \left(-\frac{4}{2}, -3\right) = (-2, -3)$. **Step 3**: Slopes of arms. - For $x \geq -2$: $2x+4 \geq 0$, so $h(x) = -(2x+4) - 3 = -2x -4 -3 = -2x -7$, slope = $-2$. - For $x < -2$: $2x+4 < 0$, so $h(x) = -(-(2x+4)) - 3 = (2x+4) - 3 = 2x +1$, slope = $2$. **Step 4**: Key points. - Vertex: $(-2, -3)$. - When $x = -1$: $h(-1) = -|2(-1)+4|-3 = -|2|-3 = -2-3=-5$ → $(-1,-5)$. - When $x = -3$: $h(-3) = -|2(-3)+4|-3 = -|-2|-3 = -2-3=-5$ → $(-3,-5)$. - When $x = 0$: $h(0) = -|0+4|-3 = -4-3=-7$ → $(0,-7)$. - When $x = -4$: $h(-4) = -|-8+4|-3 = -|-4|-3 = -4-3=-7$ → $(-4,-7)$. **Step 5**: Sketch an inverted V (opening downwards) with vertex at $(-2,-3)$, arms with slopes $\pm 2$. > **To Remember**: If $a < 0$, the graph opens downwards; if $a > 0$, it opens upwards. ### Solving Equations Graphically To solve $|x - h| = k$ graphically: - Graph $y = |x - h|$ and $y = k$ (a horizontal line). - The $x$-coordinates of intersection points are the solutions. - If $k < 0$, no solution (since absolute value is non-negative). - If $k = 0$, one solution: $x = h$. - If $k > 0$, two solutions: $x = h + k$ and $x = h - k$. #### Example 3: Solve $|x - 5| = 3$ graphically. **Step 1**: Graph $y = |x - 5|$: vertex at $(5,0)$, arms slope $\pm 1$. **Step 2**: Graph $y = 3$: horizontal line. **Step 3**: Find intersections. - Set $x - 5 = 3$ → $x = 8$. - Set $-(x - 5) = 3$ → $-x + 5 = 3$ → $-x = -2$ → $x = 2$. **Step 4**: Solutions: $x = 2$ and $x = 8$. > **To Remember**: $|x - h| = k$ has solutions $x = h \pm k$ when $k \geq 0$. #### Exercise 1: Graph $f(x) = 3|x + 1| - 2$ and state its vertex, domain, range, and axis of symmetry. **Solution**: - Rewrite: $f(x) = 3|x - (-1)| - 2$. - Vertex: $(-1, -2)$. - $a = 3 > 0$ → opens upwards. - Axis of symmetry: $x = -1$. - Domain: all real numbers, $(-\infty, \infty)$. - Range: $y \geq -2$, or $[-2, \infty)$. - Slopes of arms: $\pm 3$. - Key points: $(-1,-2)$, $(0,1)$ since $3|0+1|-2=3-2=1$, $(-2,1)$, $(1,4)$ since $3|1+1|-2=6-2=4$, $(-3,4)$. #### Exercise 2: Solve $|2x - 6| = 8$ graphically and verify algebraically. **Solution (Graphical)**: - Graph $y = |2x - 6| = 2|x - 3|$: vertex at $(3,0)$, arms slope $\pm 2$. - Graph $y = 8$: horizontal line. - Intersections: - $2x - 6 = 8$ → $2x = 14$ → $x = 7$. - $-(2x - 6) = 8$ → $-2x + 6 = 8$ → $-2x = 2$ → $x = -1$. - Solutions: $x = -1, 7$. **Verification (Algebraic)**: - Case 1: $2x - 6 \geq 0$ → $x \geq 3$: $2x - 6 = 8$ → $x = 7$ (valid). - Case 2: $2x - 6 < 0$ → $x < 3$: $-(2x - 6) = 8$ → $-2x + 6 = 8$ → $x = -1$ (valid). - Confirmed: $x = -1, 7$. --- ## II. Graphs of modulus or absolute value functions 2 ### Graphs of Absolute Value Quadratic Functions We now consider functions of the form $f(x) = |ax^2 + bx + c|$, where the quadratic expression inside the absolute value may take negative values. The graph is obtained by reflecting any portion of the parabola $y = ax^2 + bx + c$ that lies below the $x$-axis ($y < 0$) above the $x$-axis, while leaving the portions where $y \geq 0$ unchanged. This results in a graph that is always non-negative ($f(x) \geq 0$) and may have a W-shape or other complex form depending on the quadratic’s discriminant and leading coefficient. #### Key Steps to Graph $f(x) = |ax^2 + bx + c|$: 1. **Graph the quadratic** $y = ax^2 + bx + c$: find vertex, axis of symmetry, $x$- and $y$-intercepts. 2. **Determine where the quadratic is negative**: solve $ax^2 + bx + c < 0$. 3. **Reflect the negative part**: for all $x$ where $ax^2 + bx + c < 0$, plot $y = -(ax^2 + bx + c)$ instead. 4. **Combine**: the final graph is $y = ax^2 + bx + c$ where $ax^2 + bx + c \geq 0$, and $y = -(ax^2 + bx + c)$ where $ax^2 + bx + c < 0$. 5. **Note**: The graph touches the $x$-axis at the roots of the quadratic (if real), and may have local maxima at the vertex of the original quadratic if it opens downward and is reflected. > **To Remember**: The graph of $|quadratic|$ is always above or on the $x$-axis; it is symmetric only if the quadratic is symmetric about the $y$-axis (i.e., $bx = 0$). #### Example 1: Graph $f(x) = |x^2 - 4|$ **Step 1**: Graph $y = x^2 - 4$. - This is a parabola opening upwards ($a=1>0$). - Vertex: $(0, -4)$. - Axis of symmetry: $x = 0$. - $x$-intercepts: solve $x^2 - 4 = 0$ → $x = \pm 2$. - $y$-intercept: $(0, -4)$. **Step 2**: Determine where $x^2 - 4 < 0$. - $x^2 - 4 < 0$ → $x^2 < 4$ → $-2 < x < 2$. - So, the quadratic is negative between $x = -2$ and $x = 2$. **Step 3**: Reflect the negative part. - For $x \leq -2$ or $x \geq 2$: $f(x) = x^2 - 4$. - For $-2 < x < 2$: $f(x) = -(x^2 - 4) = -x^2 + 4$. **Step 4**: Analyze each piece. - Left arm ($x \leq -2$): $y = x^2 - 4$, parabola opening upwards, at $x=-2$, $y=0$; as $x \to -\infty$, $y \to \infty$. - Middle part ($-2 < x < 2$): $y = -x^2 + 4$, parabola opening downwards, vertex at $(0,4)$, at $x=\pm 2$, $y=0$. - Right arm ($x \geq 2$): $y = x^2 - 4$, same as left. **Step 5**: Key points. - $(-2, 0)$, $(2, 0)$: $x$-intercepts. - $(0, 4)$: maximum point (from reflected part). - $(-3, 5)$: $(-3)^2 - 4 = 9-4=5$. - $(3, 5)$: same. - $(-1, 3)$: $-(-1)^2 + 4 = -1+4=3$. - $(1, 3)$: same. **Step 6**: Sketch: a W-shaped graph with vertices at $(-2,0)$, $(0,4)$, $(2,0)$, symmetric about the $y$-axis. > **To Remember**: $|x^2 - a^2|$ produces a W-shape with zeros at $x = \pm a$ and a central maximum at $(0, a^2)$. #### Example 2: Graph $f(x) = | -x^2 + 6x - 8 |$ **Step 1**: Graph $y = -x^2 + 6x - 8$. - Factor: $-(x^2 - 6x + 8) = -(x-2)(x-4)$. - Opens downwards ($a=-1<0$). - Vertex: $x = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3$; $y = -(3)^2 + 6(3) - 8 = -9 + 18 - 8 = 1$. So vertex $(3,1)$. - Axis of symmetry: $x = 3$. - $x$-intercepts: solve $-x^2 + 6x - 8 = 0$ → $x^2 - 6x + 8 = 0$ → $(x-2)(x-4)=0$ → $x=2,4$. - $y$-intercept: $(0, -8)$. **Step 2**: Determine where $-x^2 + 6x - 8 < 0$. - Since the parabola opens downwards, it is negative outside the roots. - Solve $-x^2 + 6x - 8 < 0$ → multiply by $-1$ (reverse inequality): $x^2 - 6x + 8 > 0$ → $(x-2)(x-4) > 0$ → $x < 2$ or $x > 4$. - So, quadratic is negative for $x < 2$ and $x > 4$; non-negative for $2 \leq x \leq 4$. **Step 3**: Apply absolute value. - For $2 \leq x \leq 4$: $f(x) = -x^2 + 6x - 8$ (since quadratic $\geq 0$). - For $x < 2$ or $x > 4$: $f(x) = -(-x^2 + 6x - 8) = x^2 - 6x + 8$ (reflect the negative part). **Step 4**: Analyze each piece. - Middle part ($2 \leq x \leq 4$): $y = -x^2 + 6x - 8$, opens downwards, vertex at $(3,1)$, zeros at $x=2,4$. - Left arm ($x < 2$): $y = x^2 - 6x + 8$, opens upwards, vertex at $x=3$ (but only for $x<2$), at $x=2$, $y=4-12+8=0$; as $x \to -\infty$, $y \to \infty$. - Right arm ($x > 4$): $y = x^2 - 6x + 8$, opens upwards, at $x=4$, $y=16-24+8=0$; as $x \to \infty$, $y \to \infty$. **Step 5**: Key points. - $(2,0)$, $(4,0)$: $x$-intercepts. - $(3,1)$: maximum of middle part. - $(0,8)$: $0^2 - 0 + 8 = 8$ (left arm). - $(6,8)$: $36 - 36 + 8 = 8$ (right arm). - $(1,3)$: $1 - 6 + 8 = 3$ (left). - $(5,3)$: $25 - 30 + 8 = 3$ (right). **Step 6**: Sketch: a graph that looks like an "M" — two upward-opening parabolic arms on the sides and a downward-opening parabola in the middle, touching the $x$-axis at $x=2$ and $x=4$, with a peak at $(3,1)$. Symmetric about $x=3$. > **To Remember**: The graph of $|quadratic|$ has the same $x$-intercepts as the quadratic, and any local maximum of the original quadratic becomes a local minimum if it was negative and reflected, or remains a maximum if it was non-negative. ### Solving Inequalities Graphically To solve $|ax^2 + bx + c| < k$ or $|ax^2 + bx + c| > k$ graphically: - Graph $y = |ax^2 + bx + c|$. - Graph $y = k$ (horizontal line). - For $|quadratic| < k$: find where the absolute value graph lies **below** $y = k$. - For $|quadratic| > k$: find where it lies **above** $y = k$. - The solution is the set of $x$-values satisfying the condition. #### Example 3: Solve $|x^2 - 4| < 3$ graphically. **Step 1**: Recall graph of $f(x) = |x^2 - 4|$ from Example 1 in this section: W-shaped, vertex at $(0,4)$, zeros at $x=\pm 2$, arms going upwards. **Step 2**: Graph $y = 3$: horizontal line. **Step 3**: Find where $|x^2 - 4| < 3$, i.e., where the W-shape is **below** $y=3$. - The W-shape goes above $y=3$ near the center: at $x=0$, $y=4 > 3$. - Solve $|x^2 - 4| = 3$ to find boundaries. - Case 1: $x^2 - 4 = 3$ → $x^2 = 7$ → $x = \pm\sqrt{7} \approx \pm 2.645$. - Case 2: $-(x^2 - 4) = 3$ → $-x^2 + 4 = 3$ → $-x^2 = -1$ → $x^2 = 1$ → $x = \pm 1$. - So, $|x^2 - 4| = 3$ at $x = \pm 1, \pm\sqrt{7}$. - Now test intervals: - For $x < -\sqrt{7}$: e.g., $x=-3$, $|9-4|=5 > 3$ → not in solution. - For $-\sqrt{7} < x < -1$: e.g., $x=-2$, $|4-4|=0 < 3$ → in solution. - For $-1 < x < 1$: e.g., $x=0$, $|0-4|=4 > 3$ → not in solution. - For $1 < x < \sqrt{7}$: e.g., $x=2$, $|4-4|=0 < 3$ → in solution. - For $x > \sqrt{7}$: e.g., $x=3$, $|9-4|=5 > 3$ → not in solution. **Step 4**: Solution: $-\sqrt{7} < x < -1$ or $1 < x < \sqrt{7}$. > **To Remember**: Solving $|quadratic| < k$ involves finding where the reflected graph dips below $y=k$, often resulting in two separate intervals if $k$ is between 0 and the maximum of the reflected part. #### Exercise 1: Graph $f(x) = |x^2 - 2x - 3|$ and state its $x$-intercepts, $y$-intercept, and intervals where it is increasing/decreasing. **Solution**: - First, graph $y = x^2 - 2x - 3 = (x-3)(x+1)$. - Opens upwards ($a=1>0$). - Vertex: $x = -\frac{b}{2a} = \frac{2}{2} = 1$; $y = 1 - 2 - 3 = -4$. So $(1,-4)$. - $x$-intercepts: $x = -1, 3$. - $y$-intercept: $(0, -3)$. - Determine where quadratic $< 0$: between roots, $-1 < x < 3$. - Apply absolute value: - For $x \leq -1$ or $x \geq 3$: $f(x) = x^2 - 2x - 3$. - For $-1 < x < 3$: $f(x) = -(x^2 - 2x - 3) = -x^2 + 2x + 3$. - $x$-intercepts of $f(x)$: same as quadratic, $x = -1, 3$ (where it touches zero). - $y$-intercept: $f(0) = |0 - 0 - 3| = |-3| = 3$ → $(0,3)$. - Increasing/decreasing: - On $(-\infty, -1]$: $f(x) = x^2 - 2x - 3$, derivative $2x-2 < 0$ for $x<1$, but since $x<-1<1$, it’s decreasing? Wait: for $x < -1$, as $x$ increases (e.g., from $-2$ to $-1.5$), $x^2$ increases, $-2x$ decreases (becomes less negative), but net: at $x=-2$, $y=4+4-3=5$; at $x=-1.5$, $y=2.25+3-3=2.25$ → decreasing as $x$ increases? Actually, better to use vertex of each piece. - Left arm ($x \leq -1$): $y = x^2 - 2x - 3$, parabola opening upwards, vertex at $x=1$ (but this is to the right of the domain $x\leq -1$), so on $x \leq -1$, the function is decreasing as $x$ increases (since left of vertex). - At $x=-2$, $y=5$; $x=-1$, $y=0$ → as $x$ increases from $-2$ to $-1$, $y$ decreases from 5 to 0 → **decreasing** on $(-\infty, -1]$. - Middle part ($-1 < x < 3$): $y = -x^2 + 2x + 3$, opens downwards, vertex at $x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$. - Increasing on $(-1, 1)$: from $x=-1$, $y=0$ to $x=1$, $y= -1+2+3=4$. - Decreasing on $(1, 3)$: from $x=1$, $y=4$ to $x=3$, $y=-9+6+3=0$. - Right arm ($x \geq 3$): $y = x^2 - 2x - 3$, opens upwards, vertex at $x=1$ (left of domain), so increasing on $[3, \infty)$. - At $x=3$, $y=0$; $x=4$, $y=16-8-3=5$ → increasing. - Summary: - Increasing on $(-1, 1)$ and $[3, \infty)$. - Decreasing on $(-\infty, -1]$ and $(1, 3)$. #### Exercise 2: Solve $| -x^2 + 4x - 3 | \geq 2$ graphically and confirm algebraically. **Solution (Graphical)**: - First, graph $y = -x^2 + 4x - 3 = -(x^2 - 4x + 3) = -(x-1)(x-3)$. - Opens downwards ($a=-1<0$). - Vertex: $x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2$; $y = -(4) + 8 - 3 = 1$. So $(2,1)$. - $x$-intercepts: $x=1,3$. - $y$-intercept: $(0,-3)$. - Determine where quadratic $< 0$: outside roots, $x<1$ or $x>3$ (since opens down). - Apply absolute value: - For $1 \leq x \leq 3$: $f(x) = -x^2 + 4x - 3$ (quadratic $\geq 0$). - For $x < 1$ or $x > 3$: $f(x) = -(-x^2 + 4x - 3) = x^2 - 4x + 3$. - Graph $y = 2$: horizontal line. - Solve $|quadratic| = 2$ to find boundaries. - Case A (in $[1,3]$): $-x^2 + 4x - 3 = 2$ → $-x^2 + 4x - 5 = 0$ → $x^2 - 4x + 5 = 0$ → discriminant $16-20=-4<0$ → no solution. - Case B (outside): $x^2 - 4x + 3 = 2$ → $x^2 - 4x + 1 = 0$ → $x = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$. - $\sqrt{3} \approx 1.732$, so $x \approx 2 - 1.732 = 0.268$ and $x \approx 2 + 1.732 = 3.732$. - Check domains: $x=0.268 < 1$ → valid for left arm; $x=3.732 > 3$ → valid for right arm. - So, $|f(x)| = 2$ at $x = 2 - \sqrt{3}$ and $x = 2 + \sqrt{3}$. - Now, $|f(x)| \geq 2$ where the graph is **at or above** $y=2$. - Left arm ($x<1$): $f(x) = x^2 - 4x + 3$, opens upwards, vertex at $x=2$ (outside domain), so decreasing on $(-\infty,1)$? At $x=0$, $y=3$; $x=0.268$, $y=2$; as $x \to -\infty$, $y \to \infty$ → so for $x \leq 2 - \sqrt{3}$, $y \geq 2$. - Middle part ($1\leq x\leq 3$): $f(x) = -x^2 + 4x - 3$, max at $(2,1)$ < 2 → entire middle part is **below** $y=2$ → not in solution. - Right arm ($x>3$): $f(x) = x^2 - 4x + 3$, opens upwards, vertex at $x=2$ (left of domain), so increasing on $[3,\infty)$. At $x=3$, $y=0$; $x=3.732$, $y=2$; as $x \to \infty$, $y \to \infty$ → so for $x \geq 2 + \sqrt{3}$, $y \geq 2$. - Solution: $x \leq 2 - \sqrt{3}$ or $x \geq 2 + \sqrt{3}$. **Verification (Algebraic)**: Solve $| -x^2 + 4x - 3 | \geq 2$. This is equivalent to: $-x^2 + 4x - 3 \leq -2$ OR $-x^2 + 4x - 3 \geq 2$. **First inequality**: $-x^2 + 4x - 3 \leq -2$ → $-x^2 + 4x - 1 \leq 0$ → Multiply by $-1$: $x^2 - 4x + 1 \geq 0$ → Roots: $x = 2 \pm \sqrt{3}$ → Parabola opens upwards → $\geq 0$ outside roots: $x \leq 2 - \sqrt{3}$ or $x \geq 2 + \sqrt{3}$. **Second inequality**: $-x^2 + 4x - 3 \geq 2$ → $-x^2 + 4x - 5 \geq 0$ → Multiply by $-1$: $x^2 - 4x + 5 \leq 0$ → Discriminant $16-20=-4<0$, and opens upwards → always positive → no solution. Thus, solution is $x \leq 2 - \sqrt{3}$ or $x \geq 2 + \sqrt{3}$, matching graphical result. > **To Remember**: For $|quadratic| \geq k$, solve the two inequalities: quadratic $\leq -k$ OR quadratic $\geq k$, then take the union of solutions. --- ## Synthèse du chapitre - The absolute value function $|x|$ outputs the non-negative distance of $x$ from zero, defined piecewise as $x$ for $x \geq 0$ and $-x$ for $x < 0$. - Graphs of $y = a|bx + c| + d$ are V-shaped, with vertex at $\left(-\frac{c}{b}, d\right)$, axis of symmetry $x = -\frac{c}{b}$, and arm slopes $\pm \frac{a}{|b|}$ (adjusted for horizontal scaling). - The graph of $|quadratic|$ is formed by reflecting any negative portion of the quadratic $y = ax^2 + bx + c$ above the $x$-axis, resulting in graphs that are always non-negative and may exhibit W- or M-shapes. - Solving $|expression| = k$ graphically involves finding intersections with $y = k$; solving inequalities involves identifying regions where the graph lies above or below $y = k$. - Key transformation rules: vertical stretch/reflection via $a$, horizontal stretch/reflection via $b$, horizontal shift via $-\frac{c}{b}$, vertical shift via $d$. - The vertex of $a|x - h| + k$ is $(h, k)$, simplifying graphing when the expression inside the absolute value is linear in $x$. - For quadratic absolute values, the $x$-intercepts remain unchanged from the original quadratic, and the graph’s shape depends on the sign and vertex of the underlying quadratic. - Algebraic and graphical methods are complementary: graphs provide visual insight, while algebra confirms exact solutions, especially for irrational boundaries. - Mastery of piecewise reasoning is essential: always determine the sign of the expression inside the absolute value to remove the modulus correctly. ## Glossaire - **Absolute value (modulus)**: The non-negative value of a real number without regard to its sign, denoted $|x|$. - **Vertex**: The point where the graph of an absolute value function changes direction; for $y = a|x - h| + k$, it is $(h, k)$. - **Axis of symmetry**: The vertical line that divides the graph of an absolute value function into two mirror images; for $y = a|x - h| + k$, it is $x = h$. - **Piecewise function**: A function defined by different expressions over different intervals of the domain, essential for analyzing absolute value functions. - **Transformation**: A change applied to a parent function (e.g., shift, stretch, reflection) to obtain a new function; key for graphing $y = a|bx + c| + d$. - **Reflection**: A transformation that flips a graph over a line; in absolute value functions, occurs when $a < 0$ (over $x$-axis) or $b < 0$ (over $y$-axis). - **Domain**: The set of all possible input values ($x$-values) for which the function is defined; for absolute value functions, usually all real numbers. - **Range**: The set of all possible output values ($y$-values); for $y = a|x - h| + k$, it is $[k, \infty)$ if $a > 0$, or $(-\infty, k]$ if $a < 0$. - **Intercept**: A point where the graph crosses an axis; $x$-intercepts occur where $|expression| = 0$, $y$-intercept at $x = 0$. - **Inequality solution set**: The set of $x$-values satisfying an inequality involving absolute value, found graphically by comparing to horizontal lines or algebraically via case analysis. - **W-shape**: A characteristic graph form of $|quadratic|$ when the quadratic opens upwards and has two real roots, featuring two outer upward arms and a central downward peak. - **M-shape**: A characteristic graph form of $|quadratic|$ when the quadratic opens downwards and has two real roots, featuring two outer upward arms and a central upward peak (or valley, depending on reflection).