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Functions - Graphing Functions and Transformations

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Vertical and Horizontal Translations: A vertical translation y=f(x)+ky = f(x) + k shifts the graph up by kk units if k>0k > 0 and down if k<0k < 0. A horizontal translation y=f(xh)y = f(x - h) shifts the graph right by hh units if h>0h > 0 and left if h<0h < 0. Visually, these movements preserve the shape and orientation of the curve, simply sliding it across the Cartesian plane.

Vertical Stretching and Compression: The transformation y=af(x)y = a f(x) represents a vertical stretch from the xx-axis by a scale factor of aa. If a>1a > 1, the graph appears 'taller' or steeper as points move further from the xx-axis. If 0<a<10 < a < 1, the graph is compressed vertically, appearing 'flatter'. If aa is negative, the graph is also reflected across the xx-axis.

Horizontal Stretching and Compression: The transformation y=f(bx)y = f(bx) represents a horizontal stretch from the yy-axis by a scale factor of 1b\frac{1}{b}. Visually, if b>1b > 1, the graph is compressed toward the yy-axis (pushed inward), and if 0<b<10 < b < 1, the graph is stretched away from the yy-axis (pulled outward).

Reflections: The transformation y=f(x)y = -f(x) is a reflection in the xx-axis, where all positive yy-values become negative and vice versa, creating a 'mirror image' across the horizontal axis. The transformation y=f(x)y = f(-x) is a reflection in the yy-axis, mirroring the graph horizontally across the vertical axis.

Composite Transformations: When multiple transformations are applied, such as y=af(b(xh))+ky = a f(b(x - h)) + k, the order matters. Generally, horizontal transformations (inside the function) are handled first, often starting with the shift hh or factor bb, followed by vertical transformations (outside the function) like the stretch aa and finally the vertical shift kk.

Inverse Functions: The graph of the inverse function y=f1(x)y = f^{-1}(x) is a reflection of the original function y=f(x)y = f(x) across the identity line y=xy = x. Visually, every point (a,b)(a, b) on the original graph is mapped to (b,a)(b, a) on the inverse graph, effectively swapping the domain and range.

Absolute Value Transformations: The transformation y=f(x)y = |f(x)| takes any part of the graph lying below the xx-axis (y<0y < 0) and reflects it above the xx-axis, while parts already above remain unchanged. The transformation y=f(x)y = f(|x|) discards the portion of the graph where x<0x < 0 and replaces it with a reflection of the portion where x0x \geq 0 across the yy-axis, making the function even.

📐Formulae

y=f(x)+ky = f(x) + k (Vertical Translation)

y=f(xh)y = f(x - h) (Horizontal Translation)

y=af(x)y = a f(x) (Vertical Stretch, factor aa)

y=f(bx)y = f(bx) (Horizontal Stretch, factor 1b\frac{1}{b})

y=f(x)y = -f(x) (Reflection in xx-axis)

y=f(x)y = f(-x) (Reflection in yy-axis)

y=af(b(xh))+ky = a f(b(x - h)) + k (General Transformation Equation)

f(f1(x))=xf(f^{-1}(x)) = x (Identity property of Inverses)

💡Examples

Problem 1:

Given the parent function f(x)=x2f(x) = x^2, describe the sequence of transformations required to obtain the graph of g(x)=2(x+3)24g(x) = -2(x + 3)^2 - 4.

Solution:

Step 1: Identify the horizontal shift. The term (x+3)(x + 3) implies h=3h = -3, so the graph is translated 3 units to the left. Step 2: Identify the vertical stretch and reflection. The coefficient a=2a = -2 means the graph is stretched vertically by a scale factor of 22 and reflected across the xx-axis. Step 3: Identify the vertical shift. The constant k=4k = -4 means the graph is translated 4 units downward. Step 4: Final sequence: Horizontal shift left by 3, vertical stretch by factor of 2, reflection in xx-axis, and vertical shift down by 4.

Explanation:

This problem requires breaking down the transformation equation y=af(xh)+ky = a f(x-h) + k. Each constant (a,h,ka, h, k) corresponds to a specific geometric change.

Problem 2:

Find the inverse of the function f(x)=3x1x+2f(x) = \frac{3x - 1}{x + 2} and state the domain of f1(x)f^{-1}(x).

Solution:

Step 1: Replace f(x)f(x) with yy: y=3x1x+2y = \frac{3x - 1}{x + 2}. Step 2: Swap xx and yy to find the inverse: x=3y1y+2x = \frac{3y - 1}{y + 2}. Step 3: Solve for yy: x(y+2)=3y1x(y + 2) = 3y - 1 xy+2x=3y1xy + 2x = 3y - 1 xy3y=2x1xy - 3y = -2x - 1 y(x3)=(2x+1)y(x - 3) = -(2x + 1) y=(2x+1)x3=2x+13xy = \frac{-(2x + 1)}{x - 3} = \frac{2x + 1}{3 - x}. Step 4: State the inverse: f1(x)=2x+13xf^{-1}(x) = \frac{2x + 1}{3 - x}. Step 5: The domain of f1(x)f^{-1}(x) is all real numbers except where the denominator is zero: xR,x3x \in \mathbb{R}, x \neq 3.

Explanation:

The inverse is found by switching the roles of xx and yy. The domain of the inverse is equal to the range of the original function (where the horizontal asymptote of the original was y=3y=3).