Learn vertex form and transformations of parabolas. Grade 11 Math 20-1 Alberta - diploma prep.
Lesson 1.3 of Quadratic Functions & Equations in Math 20-1 — Alberta curriculum lessons.
Recall: Quadratic Functions. A quadratic function has a degree of 2 — the highest exponent of x must be exactly 2. Its graph is always a parabola.
To be quadratic, the highest exponent of x must be exactly 2 — not 1, not 3, and the variable cannot appear inside a square root or in a denominator.
Every parabola has these five properties. You should be able to identify all of them from a graph or equation.
The Five Properties.
Look at the graph above. Let's find every property of y = (x - 2)^2 - 3.
a) Vertex. Reading from the graph, the vertex is at (2, -3).
b) Direction of Opening and Axis of Symmetry. The parabola opens up because a = 1 > 0.
The axis of symmetry is the vertical line through the vertex: x = 2.
Minimum Value. Since the parabola opens up, it has a minimum. The minimum value is the y-coordinate of the vertex: -3.
c) Domain and Range. Domain: The parabola extends left and right forever.
D: \x x R\
Range: The lowest y-value is -3 (the vertex), and it goes up forever.
R: \y y ≥ -3, y R\
d) Finding the x- and y-intercepts algebraically.
Step 1 — Start with the given equation. y = (x - 2)^2 - 3
Step 2 — Replace x with 0. y = (0 - 2)^2 - 3
y = 4 - 3
y = 1
y-intercept: (0, 1)
Step 3 — Start with the given equation. y = (x - 2)^2 - 3
Step 4 — Replace y with 0. 0 = (x - 2)^2 - 3
Add 3 to both sides:
3 = (x - 2)^2
Take the square root of both sides (remember ±):
±√(3) = x - 2
Add 2 to both sides:
x = 2 ± √(3)
x-intercepts: x = 2 + √(3) and x = 2 - √(3)
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