Question in-qua

2.1 Nonlinear Functions - Polynomial end behavior and turning points
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  • 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?

A.

y=-110xx-4x+5

B.

y=-110xx-4x+52

C.

y=-110xx-5x+4

D.

y=-110xx-52x+4

In quadrant 3: The curve rises sharply to touch the x axis at point (negative 5

Hard-difficulty · SAT Math · Nonlinear Functions — Polynomial end behavior and turning points. Read the question above, select your answer, and check the full explanation below to understand exactly why the correct choice works.

Answer explanation

Choice B is correct. Each of the given choices is an equation of the form y=-110xx-amx+bn, where a , b , m , and n are positive constants. In the xy-plane, the graph of an equation of this form has x-intercepts at x = 0 , x = a , and x = - b . The graph shown has x-intercepts at x = 0 , x = 4 , and x = -5 . Therefore, a = 4 and b = 5 . Of the given choices, only choices A and B have a = 4 and b = 5 . For an equation in the form y=-110xx-amx+bn, if all values of x that are less than - b or greater than a 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 x less than -5 or greater than 4 correspond to negative y-values. Therefore, the sum of all the exponents of the factors on the right-hand side of the equation y=-110xx-4mx+5n must be even. For choice A, the sum of these exponents is 1+1+1, or 3 , which is odd. For choice B, the sum of these exponents is 1+1+2, or 4 , which is even. Therefore, y=-110xx-4x+52 could be the equation of the graph shown.

Choice A is incorrect. For the graph of this equation, all values of x less than -5 correspond to positive, not negative, y-values.

Choice C is incorrect. The graph of this equation has x-intercepts at x = -4 , x = 0 , and x = 5 , rather than x-intercepts at x = -5 , x = 0 , and x = 4 .

Choice D is incorrect. The graph of this equation has x-intercepts at x = -4 , x = 0 , and x = 5 , rather than x-intercepts at x = -5 , x = 0 , and x = 4 .