Quadratic Functions and Parabola Graphing

Introduction to Polynomials

• A polynomial function consists of terms with non-negative integer powers of the variable. It may or may not include a constant term.

• The degree of a polynomial is the highest power of the variable in the function.

• The leading coefficient is the coefficient of the term with the highest power.

• There is an infinite number of higher-order polynomials.

Quadratic Functions

• A quadratic function is a polynomial of order two (degree two).

• Standard Form: Generally written as $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are coefficients.

The Quadratic Formula

• This formula provides the solution(s) for $x$ in a quadratic equation ($ax^2 + bx + c = 0$).

• Formula: $x = rac{-b <br>ightleftharpoons rac{}{2b^2 - 4ac}}{2a}$

• The "$\pm$" symbol indicates that there are typically two possible solutions for $x$.

Parabolas: The Graph of Quadratic Functions

• When a quadratic function is graphed, it forms a curve called a parabola.

• Important Characteristics of Parabolas:

• Vertex: The lowest (minimum) or highest (maximum) point on the parabola. Its position depends on whether the parabola opens up or down.

• Y-intercept: The point where the parabola crosses the y-axis. In the standard orientation, there will always be a y-intercept.

• X-intercepts (Zeros or Roots): The point(s) where the parabola crosses the x-axis. A parabola can have two, one, or no x-intercepts, depending on its position on the Cartesian plane.

• Axis of Symmetry: A vertical line that passes through the vertex and divides the parabola into two symmetrical halves.

• Direction of Opening:

• If the leading coefficient $a$ is positive (a > 0), the parabola opens upward.

• If the leading coefficient $a$ is negative (a < 0), the parabola opens downward.

• Example Parabola Analysis:

• Vertex: $(-2, -1)$

• Axis of Symmetry: $x = -2$

• Y-intercept: $(0, 3)$

• X-intercepts: $(-1, 0)$ and $(-3, 0)$

Graphing Quadratic Functions: Vertex Form

• The Vertex Form of a quadratic function is $f(x) = a(x-h)^2 + k$

• Interpreting Coefficients:

• a:

  • Determines if the parabola opens up (a > 0) or down (a < 0).

  • Its absolute value ($|a|$) determines the vertical stretch or compression:

  • If |a| > 1, the parabola is vertically stretched (appears narrower).

  • If 0 < |a| < 1, the parabola is vertically compressed (appears wider).

• h: Represents a horizontal shift of the parabola. If $h$ is positive, the shift is to the right; if negative, to the left.

• k: Represents a vertical shift of the parabola. If $k$ is positive, the shift is upward; if negative, downward.

• Identifying Key Characteristics from Vertex Form:

• Vertex: The coordinates are $(h, k)$.

• Axis of Symmetry: The line $x = h$.

• Y-intercept: Found by evaluating $f(0)$ (setting $x=0$ and solving for $y$).

• X-intercepts: Found by setting the function $f(x)$ equal to zero and solving for $x$.

• Example 1: Graphing $g(x) = - rac{1}{2}x^2$

• This function is in vertex form ($a = - rac{1}{2}$, $h = 0$, $k = 0$).

• Vertex: $(0, 0)$.

• Axis of Symmetry: $x = 0$.

• Y-intercept: $g(0) = - rac{1}{2}(0)^2 = 0$, so $(0, 0)$.

• X-intercepts: Setting $g(x)=0 <br>ightarrow - rac{1}{2}x^2 = 0 <br>ightarrow x = 0$, so $(0, 0)$.

• Opens Down: Because $a = - rac{1}{2}$ (negative).

• Domain: $(-\&amp;, \&amp;)$.

• Range: $(-\&amp;, 0]$.

• Example 2: Graphing $F(x) = - rac{1}{2}(x-4)^2 + 3$

• This is in vertex form ($a = - rac{1}{2}$, $h = 4$, $k = 3$).

• Vertex: $(4, 3)$.

• Axis of Symmetry: $x = 4$.

• Opens Down: Because $a = - rac{1}{2}$ (negative).

• Y-intercept: $F(0) = - rac{1}{2}(0-4)^2 + 3 = - rac{1}{2}(16) + 3 = -8 + 3 = -5$, so $(0, -5)$.

• X-intercepts: Set $F(x)=0 ightarrow - rac{1}{2}(x-4)^2 + 3 = 0 ightarrow rac{1}{2}(x-4)^2 = 3 ightarrow (x-4)^2 = 6$.

  • $x-4 = \pm rac{C}{6} \implies x = 4 \pm rac{C}{6}$.

  • Solutions are approximately $(6.449, 0)$ and $(1.551, 0)$.

• Domain: $(-\&amp;, \&amp;)$.

• Range: $(-\&amp;, 3]$.

Graphing Quadratic Functions: Standard Form

• When a quadratic function is given in standard form ($f(x) = ax^2 + bx + c$), it can be converted to vertex form to identify characteristics, or specific formulas can be used.

• Method 1: Completing the Square

1. Factor out the leading coefficient $a$ from the $x^2$ and $x$ terms: $a(x^2 + rac{b}{a}x) + c$.

2. Take half of the new $x$ coefficient ($rac{b}{2a}$), square it ($( rac{b}{2a})^2$), and add/subtract it inside the parentheses. To maintain equality, subtract $a imes ( rac{b}{2a})^2$ outside the parentheses.

3. Factor the perfect square trinomial: $a(x + rac{b}{2a})^2 + (c - a( rac{b}{2a})^2)$.

4. Simplify to get the vertex form.

• Method 2: Using Formulas for Vertex Coordinates

• The x-coordinate of the vertex (h) is given by $h = - rac{b}{2a}$.

• The y-coordinate of the vertex (k) is found by evaluating the function at $x=h$: $k = f(h)$.

• Example 1: Graphing $f(x) = x^2 - 6x + 7$

• Completing the Square:

  • $x^2 - 6x + 9 - 9 + 7$

  • $(x-3)^2 - 2$

• Vertex Form: $f(x) = (x-3)^2 - 2$

• Vertex: $(3, -2)$

• Axis of Symmetry: $x = 3$

• Opens Up: (leading coefficient $a=1$ is positive).

• Y-intercept: $f(0) = (0-3)^2 - 2 = 9 - 2 = 7$, so $(0, 7)$.

• X-intercepts: Set $f(x) = 0 ightarrow (x-3)^2 - 2 = 0 ightarrow (x-3)^2 = 2$

  • $x-3 = \pm rac{C}{2} \implies x = 3 \pm rac{C}{2}$.

• Domain: $(-\&amp;, \&amp;)$.

• Range: $[-2, \&amp;)$.

• Increasing Interval: $(3, \&amp;)$.

• Decreasing Interval: $(-\&amp;, 3)$.

• Example 2: Graphing $f(x) = -3x^2 - 2x + 1$

• Completing the Square:

  • Factor out -3: $-3(x^2 + rac{2}{3}x - rac{1}{3})$.

  • Add/Subtract $( rac{1}{3})^2 = rac{1}{9}$ inside: $-3(x^2 + rac{2}{3}x + rac{1}{9} - rac{1}{9} - rac{1}{3})$.

  • Factor and simplify: $-3((x+ rac{1}{3})^2 - rac{4}{9}) = -3(x+ rac{1}{3})^2 + rac{4}{3}$.

• Vertex Form: $f(x) = -3(x+ rac{1}{3})^2 + rac{4}{3}$.

• Vertex: $(- rac{1}{3}, rac{4}{3})$.

• Axis of Symmetry: $x = - rac{1}{3}$.

• Opens Down: (leading coefficient $a=-3$ is negative).

• Y-intercept: $f(0) = -3(0)^2 - 2(0) + 1 = 1$, so $(0, 1)$.

• X-intercepts: Set $f(x)=0 ightarrow -3(x+ rac{1}{3})^2 + rac{4}{3} = 0 ightarrow (x+ rac{1}{3})^2 = rac{4}{9}$.

  • $x+ rac{1}{3} = \pm rac{2}{3}$.

  • Solutions: $x<em>1 = rac{1}{3}$ and $x</em>2 = -1$.

  • Points: $( rac{1}{3}, 0)$ and $(-1, 0)$.

• Example 3: Finding Vertex and Axis of Symmetry for $f(x) = 2x^2 + 4x + 5$

• Using Vertex Formulas:

  • $h = - rac{b}{2a} = - rac{4}{2(2)} = -1$.

  • $k = f(-1) = 2(-1)^2 + 4(-1) + 5 = 2 - 4 + 5 = 3$.

  • Vertex: $(-1, 3)$.

  • Axis of Symmetry: $x = -1$.

• Completing the Square (Alternative Method):

  • Factor out 2: $2(x^2 + 2x + rac{5}{2})$.

  • Add/Subtract $( rac{2}{2})^2 = 1$ inside: $2(x^2 + 2x + 1 - 1 + rac{5}{2})$.

  • Factor and simplify: $2((x+1)^2 + rac{3}{2}) = 2(x+1)^2 + 3$.

  • Vertex Form: $f(x) = 2(x+1)^2 + 3$.

  • Vertex: $(-1, 3)$.

  • Axis of Symmetry: $x = -1$.

Real-World Application: Rocket Trajectory Problem

• Scenario: A rocket's height ($h(t)$) at time ($t$) is modeled by the quadratic function $h(t) = -16t^2 + 128t + 80$. It's launched from $80$ feet high with an initial speed of $128$ ft/s.

• This quadratic models the parabolic trajectory of the rocket.

• Question 1: How long will it take the rocket to reach its maximum height?

• This corresponds to the time (t-coordinate) of the parabola's vertex.

• Using $t_{vertex} = - rac{b}{2a}$, where $a=-16$ and $b=128$:

  • $t = - rac{128}{2(-16)} = - rac{128}{-32} = 4$ seconds.

• The rocket reaches its maximum height after $4$ seconds.

• Question 2: What is the maximum height of the rocket?

• This corresponds to the height (h(t)-coordinate) of the parabola's vertex.

• Evaluate $h(t)$ at the vertex time ($t=4$):

  • $h(4) = -16(4)^2 + 128(4) + 80 = -16(16) + 512 + 80 = -256 + 512 + 80 = 336$ feet.

• The maximum height reached is $336$ feet.

• The vertex coordinates are $(4, 336)$.

• Question 3: How long will it take for the rocket to hit the ground?

• This corresponds to the positive x-intercept (or t-intercept) of the parabola, where $h(t) = 0$.

• Use the quadratic formula for $-16t^2 + 128t + 80 = 0$ ($a=-16, b=128, c=80$):

  • $t = rac{-128 \pm \sqrt{128^2 - 4(-16)(80)}}{2(-16)}$

  • $t = rac{-128 \pm \sqrt{16384 + 5120}}{-32}$

  • $t = rac{-128 \pm \sqrt{21504}}{-32}$

  • $rac{C}{21504} \approx 146.64$

  • $t rac{\dots}{2} \frac{-128 \pm 146.64}{-32}$

  • Two approximate solutions:

  • $t_1 = rac{-128 + 146.64}{-32} \approx -0.58$ seconds.

  • $t_2 = rac{-128 - 146.64}{-32} \approx 8.58$ seconds.

• Contextual Interpretation: Time cannot be negative, so the solution $t_1 \approx -0.58$ seconds is disregarded.

• The rocket will hit the ground approximately $8.58$ seconds after launch.

• Always consider the physical or contextual meaning of solutions in word problems.