Comprehensive Study Guide for Exponential, Quadratic, and Hyperbolic Graphs Functions

Properties and Definitions of the Exponential Function

An exponential function is a mathematical relationship where the input variable $x$ appears in the exponent of a power with a constant base. The general equation for an exponential function is given by $y = a \cdot b^x + q$. The term "mother exponential function" refers to the basic form $y = b^x$, where $b > 0$ and $b \neq 1$. Examples of mother functions include $y = 2^x$, $y = 3^x$, $y = 4^x$, $y = (\frac{1}{2})^x$, $y = (\frac{1}{3})^x$, and $y = (\frac{1}{4})^x$.

The shape of the graph depends primarily on the value of the base $b$. If $b > 1$, the graph has a "J-shape." In this case, the greater the value of $b$, the "flatter" the graph starts and the "steeper" it ends, making the J-shape look more "upright." If $0 < b < 1$, the graph has a "reverse J-shape." The closer the value of $b$ gets to $0$, the steeper the graph starts and the flatter it ends, making the reverse J-shape appear more "upright." Every mother function of the form $y = b^x$ has a y-intercept at $y = 1$ because $b^0 = 1$ for all valid values of $b$. These functions also feature a horizontal asymptote at $y = 0$, which corresponds to the x-axis.

The Equation of the Asymptote, Domain, and Range

The horizontal asymptote of an exponential function in the form $y = a \cdot b^x + q$ is defined by the equation $y = q$. This represents the vertical shift of the graph. For all exponential functions, the domain is the set of all real numbers, denoted as $x \in \mathbb{R}$.

The range of the function is determined by the sign of the constant $a$:

• If $a > 0$: Range is $y > q$ or $y \in (q; \infty)$.
• If $a < 0$: Range is $y < q$ or $y \in (-\infty; q)$.

Transformations: Stretches, Reflections, and Shifts

Multiplying the function by a constant $a$ results in vertical stretches or squashes. The greater the value of $|a|$, the more the graph is stretched away from its horizontal asymptote. For example, the steepness increases from $y = 2^x$ to $y = 2 \cdot 2^x$ to $y = 3 \cdot 2^x$. If $a < 0$, the graph is reflected about the horizontal asymptote. For instance, $y = -2^x$ is a reflection of $y = 2^x$ across the x-axis (where $q = 0$). It is important to note that $-2^x$ is equivalent to $-1 \cdot 2^x$.

The constant $q$ indicates a vertical shift. If $q > 0$, the entire graph shifts upward by $q$ units. If $q < 0$, the graph shifts downward by $q$ units. The value of $q$ also dictates the position of the horizontal asymptote.

Procedure for Sketching Exponential Graphs

To sketch a function in the form $y = a \cdot b^x + q$, follow these steps:

1. Determine the orientation: Identify if it is a J-shape ($b > 1$) or a reverse J-shape ($0 < b < 1$), and check if $a$ is negative (indicating a reflection).
2. Identify the horizontal asymptote: Draw the line $y = q$.
3. Find the y-intercept: Set $x = 0$ and solve for $y$.
4. Find the x-intercept: Set $y = 0$ and solve for $x$. This is only necessary if the graph crosses the x-axis, which occurs if the signs of $a$ and $q$ are different.
5. Determine an additional point: If more detail is needed, substitute a simple value for $x$ (such as $x = 1$ or $x = -1$) to find a corresponding $y$ value.

For example, to sketch $f(x) = -2 \cdot (\frac{1}{2})^x + 8$:

• Shape: Reverse J-shape ($b = \frac{1}{2}$), reflected and stretched ($a = -2$), shifted $8$ units up ($q = 8$).
• Horizontal Asymptote: $y = 8$.
• y-intercept: $f(0) = -2 \cdot (\frac{1}{2})^0 + 8 = -2(1) + 8 = 6$. Point: $(0; 6)$.
• x-intercept: $-2 \cdot (\frac{1}{2})^x + 8 = 0 \rightarrow -2 \cdot (2^{-1})^x = -8 \rightarrow 2^{-x} = 4 \rightarrow 2^{-x} = 2^2 \rightarrow x = -2$. Point: $(-2; 0)$.

Determining the Equation of a Specific Exponential Function

If the graph of a function $h(x) = -3 \cdot (\frac{1}{2})^x + 12$ is given, several properties can be extracted:

• Domain: $x \in \mathbb{R}$.
• Range: $y \leq 12$ or $y \in (-\infty; 12]$. (Note: There is a discrepancy in the original solution where it lists rounded vs. square brackets; for the asymptote, it is strictly $y < 12$.)
• Equation of the asymptote: $y = 12$.
• y-intercept (Point A): $y = h(0) = -3 \cdot (\frac{1}{2})^0 + 12 = 9$. Coordinates: $A(0; 9)$.
• x-intercept (Point B): $-3 \cdot (\frac{1}{2})^x + 12 = 0 \rightarrow -3 \cdot (\frac{1}{2})^x = -12 \rightarrow (\frac{1}{2})^x = 4 \rightarrow 2^{-x} = 2^2 \rightarrow x = -2$. Coordinates: $B(-2; 0)$.

In Exercise 9, find the equation for $f(x) = a \cdot 3^x + q$ given point $(2; 22)$ and asymptote $y = 4$:

• Substitute $q = 4$: $y = a \cdot 3^x + 4$.
• Substitute $(2; 22)$: $22 = a \cdot 3^2 + 4 \rightarrow 18 = 9a \rightarrow a = 2$.
• Final Equation: $f(x) = 2 \cdot 3^x + 4$.

For $g(x) = a \cdot (\frac{1}{3})^x + q$ with point $(1; -5)$ and asymptote $y = -3$:

• Substitute $q = -3$: $y = a \cdot (\frac{1}{3})^x - 3$.
• Substitute $(1; -5)$: $-5 = a \cdot (\frac{1}{3})^1 - 3 \rightarrow -2 = \frac{1}{3}a \rightarrow a = -6$.
• Final Equation: $g(x) = -6 \cdot (\frac{1}{3})^x - 3$.

Properties and Equations of Quadratic Functions (Parabolas)

A quadratic function of the form $y = ax^2 + q$ describes a parabola with its turning point on the y-axis.

• The turning point is always at the coordinates $(0; q)$.
• The axis of symmetry is the line $x = 0$ (the y-axis).
• The domain for all quadratic functions is $x \in \mathbb{R}$.
• Range if $a > 0$: $y \geq q$ or $y \in [q; \infty)$.
• Range if $a < 0$: $y \leq q$ or $y \in (-\infty; q]$.

To find the equation when the turning point $(0; q)$ and another point $(x; y)$ are known, substitute the y-coordinate of the turning point as $q$ and use the other point to solve for $a$. For example, if a parabola has a turning point $(0; -4)$ and passes through $(-1; 5)$, the calculation is: $y = ax^2 - 4$$5 = a(-1)^2 - 4$$5 = a - 4 \rightarrow a = 9$ Equation: $y = 9x^2 - 4$.

If the x-intercepts ($x_1$ and $x_2$) and another point on the graph are given, use the formula $y = a(x - x_1)(x - x_2)$. For example, with x-intercepts at $(-2; 0)$ and $(2; 0)$ and a point $(3; -15)$, the equation is found as follows: $y = a(x - (-2))(x - 2)$$y = a(x + 2)(x - 2)$$-15 = a(3 + 2)(3 - 2)$$-15 = a(5)(1) \rightarrow a = -3$$y = -3(x^2 - 4)$ Equation: $f(x) = -3x^2 + 12$.

Introduction to Hyperbolic Functions

The function $h(x) = \frac{k}{x} + q$ is a hyperbola. For the specific function $h(x) = \frac{8}{x} + 4$ shown in the diagrams:

• Domain: $x \neq 0$ or $x \in \{x \in \mathbb{R} | x \neq 0\}$.
• Range: $y \neq 4$ or $y \in \{y \in \mathbb{R} | y \neq 4\}$.
• Vertical Asymptote: $x = 0$.
• Horizontal Asymptote: $y = 4$.
• Axis of Symmetry (Positive Gradient): For a hyperbola shifted vertically by $q$, the axis of symmetry with a positive gradient is $y = x + q$. For this function, it is $y = x + 4$.
• Axis of Symmetry (Negative Gradient): The equation is $y = -x + q$. For this function, it is $y = -x + 4$.
• y-intercept: There is no y-intercept as the graph consists of two curves that never touch the vertical asymptote $x = 0$.
• x-intercept (Point A): Set $y = 0$ in $0 = \frac{8}{x} + 4 \rightarrow -4 = \frac{8}{x} \rightarrow -4x = 8 \rightarrow x = -2$. Coordinates: $A(-2; 0)$.