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how to identify the vertex, domain, range, and opening of the graph of quadratic function f(x)=x2

Sagot :

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f(x) = x² Let y = f(x), then:

y = x²

Make it in the form of (y - k) = a(x - h)², where (h, k) is the vertex.

y - 0 = 1(x - 0)²

Thus, its vertex is at (0, 0).

The domain is all possible values of x, and the range is all possible values of y.
In this equation, any value of x will suffice and not one value will make the equation invalid, thus the domain of f(x) is ALL REAL NUMBERS.
Since the a, 1, in the equation is positive, thus the parabola points upward. Thus its range is ALL REAL NUMBERS, Y IS GREATER THAN OR EQUAL TO 0

Hope you get it. Message me if you have any questions :)
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