Quadratic Functions, Forms, Graphing, Transformations, and Inequalities

Quadratic Parent Function and Fundamental Anatomy

• Definition of Quadratic Parent Function:

  • The simplest quadratic function is the parent function defined by $f(x) = x^2$.
  • Graphing the parent function creates a symmetrical, U-shaped plane curve termed a parabola ($\cup$).

• Axis of Symmetry (AOS):

  • The axis of symmetry is a vertical line that bisects the parabola into two congruent, mirror-image halves.
  • For the parent function $f(x) = x^2$, the equation for the axis of symmetry is the line $x = 0$.

• Vertex:

  • The turning point where the parabola changes direction between decreasing and increasing is the vertex.
  • For the parent function, the vertex is positioned at the origin, $V(0, 0)$.

• Standard Parent Function Table of Values:

  • $x = -3 \implies f(-3) = (-3)^2 = 9$
  • $x = -2 \implies f(-2) = (-2)^2 = 4$
  • $x = -1 \implies f(-1) = (-1)^2 = 1$
  • $x = 0 \implies f(0) = (0)^2 = 0$ (Vertex)
  • $x = 1 \implies f(1) = (1)^2 = 1$
  • $x = 2 \implies f(2) = (2)^2 = 4$
  • $x = 3 \implies f(3) = (3)^2 = 9$

• Heights from the Vertex (Step Pattern):

  • When moving horizontally by consecutive integer distances $1, 2, 3$ units from the vertex along the horizontal axis, the vertical displacements (heights) from the vertex correspond to $1^2 = 1$, $2^2 = 4$, and $3^2 = 9$.

Standard Form of Quadratic Functions: Properties and Analysis

• General Algebraic Equation:

  • Standard form is expressed as $f(x) = ax^2 + bx + c$, where $a \ne 0$, $b$, and $c$ are real coefficients.

• Parabola Orientation (Direction of Opening):

  • If the leading coefficient $a$ is positive ($a > 0$), the parabola opens upward ($\cup$).
  • If the leading coefficient $a$ is negative ($a < 0$), the parabola opens downward ($\cap$).

• Extrema (Minimum vs. Maximum):

  • When $a > 0$, the vertex represents the lowest point of the parabola and is classified as a minimum.
  • When $a < 0$, the vertex represents the highest point of the parabola and is classified as a maximum.

• Axis of Symmetry Calculation:

  • The vertical axis of symmetry is computed using the formula:     $x = \frac{-b}{2a}$
  • The numerator $-b$ represents the opposite value of $b$.

• Vertex Coordinates:

  • The vertex coordinate pair is found by evaluating the function at the axis of symmetry:     $\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$

• Comprehensive Standard Form Examples:

  • Example 1: $f(x) = 4x^2 + 0x + 0$
  • Coefficients: $a = 4$, $b = 0$, $c = 0$.
  • Direction and extremum: Since $a = 4 > 0$, the parabola features a vertical stretch by $4$ and has a minimum.
  • Axis of Symmetry: $x = \frac{-0}{2(4)} = 0$.
  • Vertex: $f(0) = 4(0)^2 = 0 \implies (0, 0)$.
  • Example 2: $f(x) = -x^2 + 6x$
  • Coefficients: $a = -1$, $b = 6$, $c = 0$.
  • Direction and extremum: Since $a = -1 < 0$, the parabola opens downward and has a maximum.
  • Axis of Symmetry: $x = \frac{-6}{2(-1)} = \frac{-6}{-2} = 3$.
  • Vertex: $f(3) = -(3)^2 + 6(3) = -9 + 18 = 9 \implies (3, 9)$.
  • Example 3: $f(x) = -2x^2 + 7$
  • Coefficients: $a = -2$, $b = 0$, $c = 7$.
  • Direction and extremum: Since $a = -2 < 0$, the parabola opens downward and has a maximum.
  • Axis of Symmetry: $x = \frac{-0}{2(-2)} = 0$.
  • Vertex: $f(0) = -2(0)^2 + 7 = 7 \implies (0, 7)$.
  • Example 4: $f(x) = x^2 - 8x + 9$
  • Coefficients: $a = 1$, $b = -8$, $c = 9$.
  • Direction and extremum: Since $a = 1 > 0$, the parabola opens upward and has a minimum.
  • Axis of Symmetry: $x = \frac{-(-8)}{2(1)} = \frac{8}{2} = 4$.
  • Vertex: $f(4) = (4)^2 - 8(4) + 9 = 16 - 32 + 9 = -7 \implies (4, -7)$.
  • Example 5: $y = x^2 - 4x + 1$
  • Coefficients: $a = 1$, $b = -4$, $c = 1$.
  • Axis of Symmetry: $x = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2$.
  • Vertex calculation: $y = (2)^2 - 4(2) + 1 = 4 - 8 + 1 = -3 \implies (2, -3)$.

Graphing Quadratic Functions by Table of Values

• Three-Step Graphing Procedure:

  • Step 1: Calculate the axis of symmetry line $x = \frac{-b}{2a}$ and evaluate the vertex coordinates $\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$.
  • Step 2: Place the vertex in the middle row of a $5$-point or $7$-point table of values. Number $x$-values symmetrically upward and downward, then evaluate each corresponding $f(x)$.
  • Step 3: Plot the ordered pairs, sketch the smooth curve with arrows, and state the domain and range.

• Using Vertical Scale Factor ($a$) for Heights from Vertex:

  • In standard parent parabolas, vertical distances from the vertex scale as $1, 4, 9$.
  • When scaled by $a$, the relative displacements become $1(a)$, $4(a)$, and $9(a)$.

• Graphing Exercises and Solutions:

  • Problem 1: $f(x) = -3x^2$
  • Parameters: $a = -3$, $b = 0$, vertical stretch by $3$, reflected downward.
  • Axis of Symmetry: $x = 0$.
  • Vertex: $(0, 0)$.
  • Table values using $a \times \text{height}$:
    • $x = -3 \implies f(-3) = 9(-3) = -27$
    • $x = -2 \implies f(-2) = 4(-3) = -12$
    • $x = -1 \implies f(-1) = 1(-3) = -3$
    • $x = 0 \implies f(0) = 0$
    • $x = 1 \implies f(1) = 1(-3) = -3$
    • $x = 2 \implies f(2) = 4(-3) = -12$
    • $x = 3 \implies f(3) = 9(-3) = -27$
  • Domain: $(-\infty, \infty)$
  • Range: $(-\infty, 0]$
  • Problem 2: $f(x) = x^2 - 6x + 11$
  • Parameters: $a = 1$, $b = -6$, $c = 11$.
  • Axis of Symmetry: $x = \frac{-(-6)}{2(1)} = 3$.
  • Vertex: $f(3) = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2 \implies (3, 2)$.
  • Table of values centered at $x = 3$:
    • $x = 1 \implies f(1) = (1)^2 - 6(1) + 11 = 6$
    • $x = 2 \implies f(2) = (2)^2 - 6(2) + 11 = 3$
    • $x = 3 \implies f(3) = 2$
    • $x = 4 \implies f(4) = (4)^2 - 6(4) + 11 = 3$
    • $x = 5 \implies f(5) = (5)^2 - 6(5) + 11 = 6$
  • Domain: $(-\infty, \infty)$
  • Range: $[2, \infty)$
  • Problem 3: $f(x) = 2x^2 - 8x + 6$
  • Parameters: $a = 2$, $b = -8$, $c = 6$.
  • Axis of Symmetry: $x = \frac{-(-8)}{2(2)} = \frac{8}{4} = 2$.
  • Vertex: $f(2) = 2(2)^2 - 8(2) + 6 = 8 - 16 + 6 = -2 \implies (2, -2)$.
  • Heights from vertex: $1(2) = 2$, $4(2) = 8$, $9(2) = 18$.
  • Table of values:
    • $x = 0 \implies f(0) = -2 + 8 = 6$
    • $x = 1 \implies f(1) = -2 + 2 = 0$
    • $x = 2 \implies f(2) = -2$
    • $x = 3 \implies f(3) = -2 + 2 = 0$
    • $x = 4 \implies f(4) = -2 + 8 = 6$
    • $x = 5 \implies f(5) = -2 + 18 = 16$
  • Domain: $(-\infty, \infty)$
  • Range: $[-2, \infty)$
  • Problem 4: $f(x) = -x^2 + 5$
  • Parameters: $a = -1$, $b = 0$, $c = 5$.
  • Axis of Symmetry: $x = 0$.
  • Vertex: $f(0) = -(0)^2 + 5 = 5 \implies (0, 5)$.
  • Domain: $(-\infty, \infty)$
  • Range: $(-\infty, 5]$
  • Problem 5: $f(x) = -\frac{1}{2}x^2 - x$
  • Parameters: $a = -\frac{1}{2}$, $b = -1$, $c = 0$.
  • Axis of Symmetry: $x = \frac{-(-1)}{2\left(-\frac{1}{2}\right)} = \frac{1}{-1} = -1$.
  • Vertex: $f(-1) = -\frac{1}{2}(-1)^2 - (-1) = -\frac{1}{2} + 1 = 0.5 \implies (-1, 0.5)$.
  • Domain: $(-\infty, \infty)$
  • Range: $(-\infty, 0.5]$

• Graphing Piecewise Function with Quadratic Sub-Function:

  • Function specification:     $f(x) = \begin{cases} 2x + 5 & \text{if } x < -4 \\ x^2 + 3x & \text{if } x \ge -4 \end{cases}$
  • Linear Piece ($x < -4$):
  • Evaluating boundary point: $f(-4) = 2(-4) + 5 = -8 + 5 = -3$.
  • Because of the strict inequality ($x < -4$), the endpoint at $(-4, -3)$ is an open circle.
  • Quadratic Piece ($x \ge -4$):
  • Evaluating boundary point: $f(-4) = (-4)^2 + 3(-4) = 16 - 12 = 4$.
  • Because the inequality is non-strict ($x \ge -4$), the endpoint at $(-4, 4)$ is a solid/closed circle.
  • Finding the quadratic vertex: $a = 1$, $b = 3$.
    • Axis of Symmetry: $x = \frac{-3}{2(1)} = -1.5$.
    • Vertex evaluation: $f(-1.5) = (-1.5)^2 + 3(-1.5) = 2.25 - 4.5 = -2.25$.
    • Vertex coordinates: $(-1.5, -2.25)$.
  • Overall Domain and Range:
  • Domain: $(-\infty, \infty)$
  • Range: $(-\infty, -3) \cup [-2.25, \infty)$

Vertex Form and Transformation Mechanics

• Algebraic Definition of Vertex Form:

  • The vertex form of a quadratic equation is defined as:     $f(x) = a(x - h)^2 + k$
  • Coordinates of the vertex are given directly by $(h, k)$.
  • Equation for the axis of symmetry is given directly by $x = h$.

• Purpose and Utility of Vertex Form:

  • Vertex form explicitly displays geometric transformations applied to the parent function $f(x) = x^2$.

• Role of Parameters:

  • Parameter $a$ (Dilation and Reflection):
  • Determines the width (dilation factor) and vertical direction of opening.
  • If $|a| > 1$, the graph undergoes a vertical stretch.
  • If $0 < |a| < 1$, the graph undergoes a vertical compression.
  • If $a < 0$, the graph is reflected across the $x$-axis.
  • Parameter $h$ (Horizontal Translation):
  • Represents a horizontal shift.
  • An expression $(x - h)^2$ shifts the graph right by $h$ units.
  • An expression $(x + h)^2 = (x - (-h))^2$ shifts the graph left by $h$ units.
  • Parameter $k$ (Vertical Translation):
  • Represents a vertical shift.
  • $+k$ translates the graph up by $k$ units.
  • $-k$ translates the graph down by $k$ units.

• Transformation Analysis of Selected Functions:

  • 1. $f(x) = x^2 + 9$: Vertical shift up 9 units.
  • 2. $f(x) = x^2 - 2$: Vertical shift down 2 units.
  • 3. $f(x) = -(x + 5)^2$: Reflection across the $x$-axis and horizontal shift left 5 units.
  • 4. $f(x) = (x - 4)^2 - 1$: Horizontal shift right 4 units and vertical shift down 1 unit.
  • 5. $f(x) = \frac{1}{3}x^2 + 2$: Vertical compression by a factor of $\frac{1}{3}$ and vertical shift up 2 units.
  • 6. $f(x) = -2(x + 4)^2$: Vertical stretch by a factor of 2, reflection across the $x$-axis, and horizontal shift left 4 units.

• Mason Transformation Word Problem:

  • Scenario: The parent quadratic function is reflected over the $x$-axis, then translated $7$ units up and $3$ units to the right.
  • Resulting Equation: $f(x) = -(x - 3)^2 + 7$
  • Resulting Vertex: $(3, 7)$
  • Resulting Axis of Symmetry: $x = 3$

• Graphing from Vertex Form:

  • Example A: $y = -2(x + 1)^2 + 6$
  • Rewrite into standard vertex form: $y = -2(x - (-1))^2 + 6$.
  • Vertex: $(-1, 6)$.
  • Axis of Symmetry: $x = -1$.
  • Orientation: Opens downward ($a = -2$).
  • Example B: $f(x) = 1(x - 2)^2 + 3$
  • Transformations: Shifted right $2$ units, shifted up $3$ units.
  • Axis of Symmetry: $x = 2$.
  • Vertex: $(2, 3)$.
  • Example C: $f(x) = 1x^2 - 4$
  • Transformations: Shifted down $4$ units.
  • Axis of Symmetry: $x = 0$.
  • Vertex: $(0, -4)$.
  • Domain: $(-\infty, \infty)$
  • Range: $[-4, \infty)$

Graphing Quadratic Inequalities

• Three-Step Procedure for Graphing Quadratic Inequalities:

  • Step 1: Create a Table of Values:
  • Locate the vertex and position it in the middle row of the table, generating symmetric points on both sides.
  • Step 2: Determine Boundary Line Style:
  • Draw a dashed line for strict inequality symbols ($<$, $>$) to denote that points on the boundary are not solutions.
  • Draw a solid line for inclusive inequality symbols ($\le$, $\ge$) to denote that points on the boundary are solutions.
  • Step 3: Determine the Shaded Solution Region (with $y$ on the left side):
  • Shade above the boundary curve for greater-than symbols ($>$ or $\ge$).
  • Shade below the boundary curve for less-than symbols ($<$ or $\le$).

• Spatial Orientation Conventions (Inside vs. Outside):

  • Parabolas opening upward ($\cup$):
  • Greater-than ($\ge$, $>$): Shading above corresponds to the inside of the parabola.
  • Less-than ($\le$, $<$): Shading below corresponds to the outside of the parabola.
  • Parabolas opening downward ($\cap$):
  • Greater-than ($\ge$, $>$): Shading above corresponds to the outside of the parabola.
  • Less-than ($\le$, $<$): Shading below corresponds to the inside of the parabola.

• Worked Examples of Quadratic Inequalities:

  • Inequality 1: $y > x^2 + 2x - 2$
  • Vertex calculation: $x = \frac{-2}{2(1)} = -1$.
  • $f(-1) = (-1)^2 + 2(-1) - 2 = 1 - 2 - 2 = -3 \implies V(-1, -3)$.
  • Boundary line: Dashed line ($>$).
  • Shading: Shaded above the curve (inside the upward-opening parabola).
  • Inequality 2: $y \le -x^2 + 4x - 1$
  • Vertex calculation: $x = \frac{-4}{2(-1)} = 2$.
  • $f(2) = -(2)^2 + 4(2) - 1 = -4 + 8 - 1 = 3 \implies V(2, 3)$.
  • Boundary line: Solid line ($\le$).
  • Shading: Shaded below the curve (inside the downward-opening parabola).
  • Inequality 3: $y \le -3x^2 + 18x - 22$
  • Vertex calculation: $x = \frac{-18}{2(-3)} = 3$.
  • $f(3) = -3(3)^2 + 18(3) - 22 = -27 + 54 - 22 = 5 \implies V(3, 5)$.
  • Boundary line: Solid line ($\le$).
  • Shading: Shaded below the curve (inside the downward-opening parabola).
  • Inequality 4: $y < x^2 + 6$
  • Vertex calculation: $x = 0 \implies V(0, 6)$.
  • Boundary line: Dashed line ($<$).
  • Shading: Shaded below the curve (outside the upward-opening parabola).
  • Inequality 5: $y \le -2x^2 - 2x$
  • Vertex calculation: $x = \frac{-(-2)}{2(-2)} = \frac{2}{-4} = -0.5$.
  • Boundary line: Solid line ($\le$).
  • Shading: Shaded below the curve (inside the downward-opening parabola).
  • Inequality 6: $y > \frac{1}{2}x^2 - 4x + 12$
  • Vertex calculation: $x = \frac{-(-4)}{2\left(\frac{1}{2}\right)} = \frac{4}{1} = 4$.
  • Boundary line: Dashed line ($>$).
  • Shading: Shaded above the curve (inside the upward-opening parabola).