1-1 Graphing Quadratic Functions Study Notes

Standard Form of a Quadratic Function
- A quadratic function is generally expressed as $f(x) = ax^2 + bx + c$ .
- The variables involved represent the following:
- $a$ : The coefficient of the quadratic term ( $x^2$ ). If $a > 0$ , the parabola opens upward, and the vertex is a minimum. If $a < 0$ , the parabola opens downward, and the vertex is a maximum.
- $b$ : The coefficient of the linear term ( $x$ ).
- $c$ : The constant term, which represents the $y$ -intercept of the function.
Determining Key Components for Graphing
- $y$ -intercept: The value of the function when $x = 0$ , which is always equal to $c$ .
- Axis of Symmetry: The vertical line that divides the parabola into two symmetric halves. The equation is calculated using the formula: $x = -\frac{b}{2a}$
- Vertex: The highest or lowest point on the parabola. Its $x$ -coordinate is the value of the axis of symmetry. To find the $y$ -coordinate, substitute the $x$ -coordinate back into the original function $f(x)$ .

Case 1: $f(x) = x^2 + 6x + 8$
- Values: $a = 1$ , $b = 6$ , $c = 8$ .
- $y$ -intercept: $8$ .
- Axis of Symmetry: $x = -\frac{6}{2(1)} = -3$ .
- Vertex: $(-3, -1)$ . Since $a > 0$ , this is a minimum.
- Table of Values for Graphing:
- $f(-5) = (-5)^2 + 6(-5) + 8 = 3$
- $f(-4) = (-4)^2 + 6(-4) + 8 = 0$
- $f(-3) = -1$
- $f(-2) = (-2)^2 + 6(-2) + 8 = 0$
- $f(-1) = (-1)^2 + 6(-1) + 8 = 3$
- Domain: All real numbers.
- Range: $\{y | y \geq -1\}$ .
Case 2: $f(x) = -x^2 - 2x + 2$
- Values: $a = -1$ , $b = -2$ , $c = 2$ .
- $y$ -intercept: $2$ .
- Axis of Symmetry: $x = -\frac{-2}{2(-1)} = -1$ .
- Vertex: $(-1, 3)$ . Since $a < 0$ , this is a maximum.
- Domain: All real numbers.
- Range: $\{y | y \leq 3\}$ .
Case 3: $f(x) = 2x^2 - 4x + 3$
- Values: $a = 2$ , $b = -4$ , $c = 3$ .
- Axis of Symmetry: $x = 1$ .
- Vertex: $(1, 1)$ . Since $a > 0$ , this is a minimum.
- Domain: All real numbers.
- Range: $\{y | y \geq 1\}$ .
Case 4: $f(x) = -2x^2$
- Values: $a = -2$ , $b = 0$ , $c = 0$ .
- Axis of Symmetry: $x = 0$ .
- Vertex: $(0, 0)$ . Since $a < 0$ , this is a maximum.
- Range: $\{y | y \leq 0\}$ .
Case 5: $f(x) = x^2 - 4x + 4$
- Values: $a = 1$ , $b = -4$ , $c = 4$ .
- Axis of Symmetry: $x = 2$ .
- Vertex: $(2, 0)$ . Since $a > 0$ , this is a minimum.
- Range: $\{y | y \geq 0\}$ .
Case 6: $f(x) = x^2 - 6x + 8$
- Values: $a = 1$ , $b = -6$ , $c = 8$ .
- Axis of Symmetry: $x = 3$ .
- Vertex: $(3, -1)$ , min.
- Range: $\{y | y \geq -1\}$ .

Vertex Comparison Examples
- Problem 7: Compare $f(x)$ (with max $-1$ ) to $g(x)$ (y-intercept $1$ , vertex at $(1, 3)$ ). $g(x)$ has the greater maximum because its vertex height ( $y = 3$ ) is $4$ units above the vertex of $f(x)$ at $(1, -1)$ .
- Problem 8: Compare $f(x)$ (min $-1$ ) to $g(x)$ (y-intercept $0.5$ , vertex at $(-1, -5)$ ). $g(x)$ has the lesser minimum because its vertex ( $y = -5$ ) is $4$ units below the vertex of $f(x)$ at $(-1, -1)$ .
- Problem 9: Compare $f(x) = x^2 - 10x + 5$ to table-defined $g(x)$ .
- $f(x)$ vertex: Axis is $x = 5$ , vertex is $(5, -20)$ , min.
- $g(x)$ vertex: Using table symmetry, vertex is $(5, -10)$ , min.
- Result: $f(x)$ has the lesser minimum ( $-20 < -10$ ).
- Problem 10: Compare $f(x) = -x^2 + 6x - 15$ to table-defined $g(x)$ .
- $f(x)$ vertex: Axis is $x = 3$ , vertex is $(3, -6)$ , max.
- $g(x)$ vertex: Middle value $(3, 1)$ , max.
- Result: $g(x)$ has the greater maximum ( $1 > -6$ ).

Revenue Optimization (Fishing Permits)
- Scenario: Last year, $480$ permits sold at $\$80$ each. For every $\$4$ increase in price, $16$ fewer permits are sold.
- Let $x$ = number of price increases; $y$ = revenue.
- Function: $P(x) = (80 + 4x)(480 - 16x) = -64x^2 + 640x + 38,400$ .
- Maximization: Axis of symmetry $x = -\frac{640}{2(-64)} = 5$ .
- Optimal Price: $80 + 4(5) = \$100$ total.
- Domain: $\{x | 0 \leq x \leq 30\}$ (permits sold cannot be negative).
- Range: $\{y | 0 \leq y \leq 40,000\}$ .
Profit Maximization (Jar Candles)
- Scenario: $120$ candles sold at $\$30$ each. For every $\$2$ decrease, $10$ additional candles sold.
- Function: $P(x) = (30 - 2x)(120 + 10x) = -20x^2 + 60x + 3600$ .
- Axis of Symmetry: $x = 1.5$ .
- Optimal Price: $30 - 2(1.5) = \$27$ .
- Domain/Range: $D = \{x | 0 \leq x \leq 15\}$ , $R = \{y | 0 \leq y \leq 3645\}$ .
Revenue (Cinema Tickets)
- Scenario: $200$ tickets at $\$6.00$ . For every $\$0.50$ increase, $10$ fewer sales.
- Function: $R(x) = -5x^2 + 40x + 1200$ .
- Optimal x: $4$ . Price per ticket: $\$6.00 + 4(0.50) = \$8.00$ .
- Max Revenue: $R(4) = \$1280.00$ .
Projectile Motion
- Ball tossed upward: $h(t) = -16t^2 + 32t + 4$ .
- Max height: $20\,\text{ft}$ reached at $t = 1\,\text{second}$ .
- Cannonball trajectory: Path described by a quadratic. If the cannon sits on the $y$ -axis and $c = 20$ , the cannon is $20\,\text{ft}$ above the ground.

Formula for Average Rate of Change
- Over the interval $[x_1, x_2]$ , the rate is defined as: $\text{Rate} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$
Specific Results (Interval-Based)
- Problem 13: $f(x) = x^2 - 10x + 5$ , interval $[-4, 4]$ . Average Rate of Change = $-10$ .
- Problem 14: $f(x) = 2x^2 + 4x - 6$ , interval $[-3, 3]$ . Average Rate of Change = $4$ .
- Problem 15: $f(x) = 3x^2 - 3x + 1$ , interval $[-5, 5]$ . Average Rate of Change = $-3$ .
- Problem 16: $f(x) = 4x^2 + x + 3$ , interval $[-2, 2]$ . Average Rate of Change = $1$ .
- Problem 17: $f(x) = 2x^2 - 11$ , interval $[-3, 3]$ . Average Rate of Change = $0$ .
- Problem 23: Table-defined function with $f(3) = 27$ and $f(-3) = 27$ . ARC over $[-3, 3] = 0$ .
- Problem 24: Table-defined function with $f(2) = -4$ and $f(-2) = 12$ . ARC over $[-2, 2] = \frac{-4 - 12}{4} = -4$ .

Quadratic Recognition
- A function is quadratic if it has no terms higher than $x^2$ , a linear term, and a constant term.
- Concavity Check:
- Leading coefficient ( $a$ ) is negative: Function has a maximum.
- Leading coefficient ( $a$ ) is positive: Function has a minimum.
Creation Examples
- Maximum of $8$ : Sample function $f(x) = -x^2 + 8$ ( $a = -1$ , axis at $x = 0$ ).
- Minimum of $-4$ : Sample function $f(x) = x^2 - 4$ ( $a = 1$ , axis at $x = 0$ ).
- Vertex of $(-2, 6)$ : Sample function $f(x) = x^2 + 4x + 10$ .
- Explanation: Using axis formula $x = -2$ , we can set $a = 1$ and solve for $b$ : $-2 = -\frac{b}{2(1)} \rightarrow b = 4$ . Then solve for $c$ to ensure the vertex $y$ -value is $6$ .