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Nonlinear functions SAT question

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

  • In quadrant 3:
    • The curve rises sharply to touch the x axis at point (negative 5 comma 0).
    • The curve falls gradually to a relative minimum at point (negative 2 comma negative 11).
    • The curve rises gradually to cross both axes at the origin.
  • In quadrant 1:
    • The curve rises gradually to a relative maximum at point (2.5 comma 21).
    • The curve falls sharply to cross the x axis at point (4 comma 0).
  • In quadrant 4 the curve falls sharply.

Which of the following could be the equation of the graph shown in the xy-plane?

ExplanationShow

Choice B is correct. Each of the given choices is an equation of the form , where , , , and are positive constants. In the xy-plane, the graph of an equation of this form has x-intercepts at , , and . The graph shown has x-intercepts at , , and . Therefore, and . Of the given choices, only choices A and B have and . For an equation in the form , if all values of that are less than or greater than correspond to negative y-values, then the sum of all the exponents of the factors on the right-hand side of the equation is even. In the graph shown, all values of less than or greater than correspond to negative y-values. Therefore, the sum of all the exponents of the factors on the right-hand side of the equation  must be even. For choice A, the sum of these exponents is , or , which is odd. For choice B, the sum of these exponents is , or , which is even. Therefore,  could be the equation of the graph shown.

Choice A is incorrect. For the graph of this equation, all values of less than correspond to positive, not negative, y-values.

Choice C is incorrect. The graph of this equation has x-intercepts at , , and , rather than x-intercepts at , , and .

Choice D is incorrect. The graph of this equation has x-intercepts at , , and , rather than x-intercepts at , , and .

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