IB Mathematics AI SL Determine key features of graphs MAI Study Notes - New Syllabus

Key Features of Graphs 

 Maximum and Minimum Values
Definition:
Maximum: The highest point (y-value) of a function within a given domain.
Minimum: The lowest point (y-value) of a function within a given domain.
Types:
Global (Absolute): The highest/lowest point over the entire domain.
Local (Relative): The highest/lowest point within a small interval (but not necessarily the entire graph).
Example:
For \( f(x) = -x^2 + 4x + 5 \), the vertex (highest point) is at (2, 9).

Intercepts (x and y)
X-intercepts (Roots):
 Where \( y = 0 \) (graph crosses the x-axis).
 Found by solving \( f(x) = 0 \).
Y-intercepts:
 Where \( x = 0 \) (graph crosses the y-axis).
 Found by evaluating \( f(0) \).
Example:
For \( f(x) = x^2 – 4x + 3 \):
x-intercepts: \( (1, 0) \) and \( (3, 0) \) (solutions to \( x^2 – 4x + 3 = 0 \)).
y-intercept: \( (0, 3) \) (since \( f(0) = 3 \)).

Symmetry
Even Functions: Symmetric about the y-axis (\( f(-x) = f(x) \)).
Example: \( f(x) = x^2 \).
Odd Functions: Symmetric about the origin (\( f(-x) = -f(x) \)).
Example: \( f(x) = x^3 \).

Asymptotes (Vertical & Horizontal)
Vertical Asymptotes:
 Occur where a function is undefined (e.g., denominator = 0).
 Example: \( f(x) = \frac{1}{x-3} \) has a vertical asymptote at \( x = 3 \).
Horizontal Asymptotes:
Describe end behavior as \( x \rightarrow \pm \infty \).
 Example: \( f(x) = \frac{2x}{x+1} \) approaches \( y = 2 \).

Using Graphing Technology
Key Features to Check:
Zeros (Roots): Where \( f(x) = 0 \).
Extrema: Max/min points.
Intersections: Points where two graphs meet.
Example:
To find where \( y = x^2 \) and \( y = 2x + 1 \) intersect:
Graph both functions.
 Use the “Intersect” tool to find exact coordinates.

 Effects of Parameters (Transformations)
General Form: \( f(x) = a(x – h)^2 + k \) (for quadratics).
Impact of Parameters:
 a: Controls width and direction (up/down).
h: Shifts graph left/right.
 k: Shifts graph up/down. 

For intersection points of \( f \) and \( g \):

\( \begin{aligned}
f(x) = g(x) &\Leftrightarrow (x – 3)^2 – 4 = x – 5 \Leftrightarrow x^2 – 6x + 9 – 4 = x – 5 \Leftrightarrow x^2 – 7x + 10 = 0 \\
&\Leftrightarrow x = 2 \text{ or } x = 5
\end{aligned} \)

By using either \( f(x) \) or \( g(x) \) we find \( y = -3 \), \( y = 0 \) respectively.
Hence, the curves intersect at points \( (2, -3) \) and \( (5, 0) \)

Indeed, the graphs of \( f(x) \) and \( g(x) \) are as follows

Remark: Confirm all the results by using GDC – Graph mode.

SOLVING EQUATIONS AND INEQUALITIES BY USING GRAPHS

We can solve

 equations of the form \( f(x) = g(x) \)
 inequalities of the form \( f(x) > g(x) \) or \( f(x) \geq g(x) \)

by using GDC – graph mode
METHOD A: we find the intersection points of the graphs

\( \begin{aligned}
y_1 &= f(x) \\
y_2 &= g(x)
\end{aligned} \)

Solutions of \( f(x) = g(x) \): \( x \)-coordinates of intersection points
Solutions of \( f(x) > g(x) \): intervals where \( y_1 = f(x) \) is above \( y_2 = g(x) \)

METHOD B: we find the roots of the graph

\( y_1 = f(x) – g(x) \)

Solutions of \( f(x) – g(x) = 0 \): the roots of the graph
Solutions of \( f(x) – g(x) > 0 \): intervals where \( y_1 = f(x) – g(x) \) is positive