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Given y = ax², what is the shape of the graph when
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
Given y = 3x², what will happen to the graph when the coefficient of x² becomes 1?
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
In standard form, a _______ function is written as y = ax² + bx + c
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
Given y = 2x², the equation for the line of symmetry for this graph is
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
This curve, y=x² - 5x + 7, has what kind of roots?
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
Given y = 2(x-1)² +1, what is kind of graph is this?
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
Given y = 2(x + 4)² -1, the coordinates of the turning point are
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
What is the general shape of parabolas?
An inverted u-shaped curve
U-shape
parabolic graph
imaginary roots
( - 4, - 1)
The graph will open outward.
quadratic
x = 0
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