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Nonlinear functions SAT question

Math · Advanced Math · Medium · Answer it here, explanation included.

$f of x equals, open parenthesis, x plus 4, close parenthesis, times, open parenthesis, x minus 1, close parenthesis, times, open parenthesis, 2 x minus 3, close parenthesis$

The function f is defined above. Which of the following is NOT an x-intercept of the graph of the function in the xy-plane?

ExplanationShow

Choice B is correct. The graph of the function f in the xy-plane has x-intercepts at the points $with coordinates x comma y$ , where $y equals f of x, which equals 0$ . Substituting 0 for $f of x$ in the given equation yields $0 equals, open parenthesis, x plus 4, close parenthesis, times, open parenthesis, x minus 1, close parenthesis, times, open parenthesis, 2 x minus 3, close parenthesis$ . By the zero product property, if $0 equals, open parenthesis, x plus 4, close parenthesis, times, open parenthesis, x minus 1, close parenthesis, times, open parenthesis, 2 x minus 3, close parenthesis$ , then $x plus 4, equals 0$ , $x minus 1, equals 0$ , or $2 x minus 3, equals 0$ . Solving each of these linear equations for x, it follows that $x equals negative 4$ , $x equals 1$ , and $x equals three halves$ , respectively. This means that the graph of the function f in the xy-plane has three x-intercepts: $the point with coordinates negative 4 comma 0$ , $the point with coordinates 1 comma 0$ , and $the point with coordinates three halves comma 0$ . Therefore, $the point with coordinates negative two thirds comma 0$ 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 $k comma 0$ 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 $negative two thirds$ for x and 0 for $f of x$ in the given equation yields $0 equals, open parenthesis, negative two thirds plus 4, close parenthesis, times, open parenthesis, negative two thirds minus 1, close parenthesis, times, open parenthesis, 2 times negative two thirds, minus 3, close parenthesis$ , or $0 equals, the fraction 650 over 27$ . Therefore, the point $with coordinates negative two thirds comma 0$ 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 $k comma 0$ , where k is a constant. Substituting $negative 4$ for x and 0 for $f of x$ in the given equation yields $0 equals, open parenthesis, negative 4 plus 4, close parenthesis, times, open parenthesis, negative 4 minus 1, close parenthesis, times, open parenthesis, 2 times negative 4, minus 3, close parenthesis$ , or $0 equals 0$ . Since this is a true statement, the point $with coordinates negative 4 comma 0$ lies on the graph of the function f and is an x-intercept of the graph. Performing similar substitution using the points $with coordinates 1 comma 0$ and $three halves comma 0$ also yields the true statement $0 equals 0$ , illustrating that these points also lie on the graph of the function f and are x-intercepts of the graph.

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