MCR3U
Ontario
11th
f(x) = x
x
y
-2
-1
0
1
2
Domain: {x E R}
Range: {y E R}
f(x) = x²
4
Range: {y E R/0 ≤ y}
f(x) = √x
Domain: {x E R/0 ≤ x}
f(x) = 1/x
-1/2
-0.5
0.5
1/2
Domain: {x E R/x ≠ 0}
Range: {y E R/y ≠ 0}
Asymptote: x = 0, y = 0
f(x) = |x|
f(x) = x³
-8
8
Transformed functions: f(x) = a(k(x-d)) + c
Vertical Stretch: a
By a factor of….
If negative, reflection in the x axis
Horizontal Stretch: k
Always 1/k (flipped)
If negative, reflection in the y axis
Vertical Translation: c
if positive, moves up
If negative, moves down
Horizontal Translation: d
Always the opposite sign of what it is in the brackets (sign is flipped)
If positive in bracket (so negative alone), then it moves left ( <-- )
If negative in bracket (so positive alone), then it moves right ( --> )
Draw the parent functions’ table of values
Create mapping notation using:
Mapping Notation: ((1/k)x + d, ay + c)
Apply mapping notation to the parent function and graph (following BEDMAS, order of operations)
Quadratic
g(x) = a(k(x-d))² + c
Reciprocal
g(x) = a(1/(k(x-d)) + c
Cubic
g(x) = a(k(x-d))³ + c
Square Root
g(x) = a(√k(x-d) ) + c
Absolute Value
g(x) = a |k(x-d)| + c
Linear:
D: {x E R}
R: {y ER}
Cubic:
Quadratic:
R: {y E R/0 ≤ y}
c is the restriction (replacing zero)
Absolute Value:
Reciprocal:
D: {x E R/ x ≠ 0}
c replaces the restriction
R: {y E R/ y ≠ 0}
d replaces the restriction
The function cannot touch the asymptote thus the asymptote is our restriction
Square Root:
D: {x E R / 0 ≤ x}
Make the number under the square root sign as small as it can be, so zero (because it cannot be negative since you can’t have a negative radicand)
R: {y E R / 0 ≤ y}
Look at what your lowest y could be as a result of the reduction of x for domain
Write the function in x-y notation
Swap x and y
Solve for y