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Through which point does the graph of f(x)=x³ pass?

Answer and explanation

Correct answer: (0,0)

The governing point condition for a function graph is that a point (x,y) lies on y=f(x) exactly when y=f(x). Test the coordinates in the options. For x=0, f(0)=0³=0, so (0,0) lies on the graph and option C is correct. For (0,1), the output would need to be 1, but f(0)=0. For (1,0), the function gives f(1)=1, not 0. For x=−1, f(−1)=(−1)³=−1, so the corresponding graph point is (−1,−1), not (−1,1). Therefore the cubic graph passes through the origin. This conclusion follows directly by substitution and does not require plotting the entire curve. The odd symmetry of x³ also agrees with the fact that the origin is on the graph.

Related tags

Relations-FunctionsCubic-FunctionStandard-GraphsPoint-On-GraphGraphs Of Standard FunctionsRelations And FunctionsMathematicsClass 11 Mcq

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