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Given y = ax², what is the shape of the graph when
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
Given y = 2(x-1)² +1, what is kind of graph is this?
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
Given y = 2x², the equation for the line of symmetry for this graph is
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
Given y = 3x², what will happen to the graph when the coefficient of x² becomes 1?
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
Given y = 2(x-1)² +1, the equation for its line of symmetry is
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
In standard form, a _______ function is written as y = ax² + bx + c
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
This curve, y=x² - 5x + 7, has what kind of roots?
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
What is the general shape of parabolas?
imaginary roots
quadratic
An inverted u-shaped curve
x = 1
The graph will open outward.
x = 0
parabolic graph
U-shape
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