Write the equation of a circle in standard form with a diameter whose endpoints are
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Let's plug in a few points to see which graph belongs to the function f(x) = 2x^2
Let's say x =0, let's find y!
2(0)^2 = 2(0) = 0
So one point on our line is (0.0)
Just finding this point right here eliminates 3 of the answer choices. Your answer is the graph that goes through the point (0,0)
For your added convenience, I have attached the graph. :)
Let's say x =0, let's find y!
2(0)^2 = 2(0) = 0
So one point on our line is (0.0)
Just finding this point right here eliminates 3 of the answer choices. Your answer is the graph that goes through the point (0,0)
For your added convenience, I have attached the graph. :)