sat-dog

Choice B is correct. The graph of the function f in the xy-plane has x-intercepts at the points , where . Substituting 0 for in the given equation yields . By the zero product property, if , then , , or . Solving each of these linear equations for x, it follows that , , and , respectively. This means that the graph of the function f in the xy-plane has three x-intercepts: , , and . Therefore, isn’t an x-intercept of the graph of the function f.
Alternate approach: Substitution may be used. Since by definition an x-intercept of any graph is a point in the form where k is a constant, and since all points in the options are in this form, it need only be checked whether the points in the options lie on the graph of the function f. Substituting for x and 0 for in the given equation yields , or . Therefore, the point doesn’t lie on the graph of the function f and can’t be an x-intercept of the graph.

Choices A, C, and D are incorrect because each of these points is an x-intercept of the graph of the function f in the xy-plane. By definition, an x-intercept is a point on the graph of the form , where k is a constant. Substituting for x and 0 for in the given equation yields , or . Since this is a true statement, the point lies on the graph of the function f and is an x-intercept of the graph. Performing similar substitution using the points and also yields the true statement , illustrating that these points also lie on the graph of the function f and are x-intercepts of the graph.