Use the image to answer the question.

A curve with 2 turning points is plotted on a coordinate grid and is labeled m left parenthesis x right parenthesis. The plot starts in quadrant 3, rises to a turning point in quadrant 1, drops to a turning point in quadrant 4 and rises through quadrant 1.

Which of the following could be the equation of m(x) in factored form, given its graph?

(1 point)
Responses

m(x)=(x+1)(x−3)(x−5)
m left parenthesis x right parenthesis equals left parenthesis x plus 1 right parenthesis left parenthesis x minus 3 right parenthesis left parenthesis x minus 5 right parenthesis

m(x)=(x−1)(x+3)(x+5)(x+15)
m left parenthesis x right parenthesis equals left parenthesis x minus 1 right parenthesis left parenthesis x plus 3 right parenthesis left parenthesis x plus 5 right parenthesis left parenthesis x plus 15 right parenthesis

m(x)=(x−1)(x+3)(x+5)
m left parenthesis x right parenthesis equals left parenthesis x minus 1 right parenthesis left parenthesis x plus 3 right parenthesis left parenthesis x plus 5 right parenthesis

m(x)=(x+1)(x−3)(x−5)(x−15)

1 answer

To determine which equation could represent the curve \( m(x) \) based on the given characteristics of the graph, we should analyze the implications of the turning points and behavior of the function.

  1. Turning Points: A turning point occurs at a root of the polynomial where the function changes direction. Given that the graph has 2 turning points, it suggests that the function can only have 2 real roots or repeated roots.

  2. End Behavior: The function starts in quadrant 3 (negative \( y \) values) and rises to a turning point in quadrant 1, then drops to another turning point in quadrant 4 (again negative \( y \) values) before rising through quadrant 1, which indicates that the leading coefficient is positive and the degree of the polynomial is odd.

Now, let’s analyze the factored forms provided:

  1. m(x) = (x+1)(x-3)(x-5)

    • This has 3 roots and thus would represent a degree 3 polynomial. It could potentially have either 1 or 3 turning points, which does not match the given requirement.
  2. m(x) = (x−1)(x+3)(x+5)(x+15)

    • This has 4 roots and corresponds to a degree 4 polynomial. It can have up to 4 turning points. This exceeds the 2 specified turning points.
  3. m(x) = (x−1)(x+3)(x+5)

    • This has 3 roots, similar to the first option and thus also does not meet the requirements.
  4. m(x) = (x+1)(x−3)(x−5)(x−15)

    • This equation has 4 roots which aligns with a degree 4 polynomial — it could potentially have multiple turning points as well.

Given that we need an equation that can produce exactly two turning points, we must pick among those options.

Upon inspection, the most appropriate choice, considering the graph's behavior and turning points, is to have at least one of the roots as repeated (which is not given in the options listed). However, since we are limited by the options given and both equations with 4 roots cannot fit the requirement and the ones with 3 roots give us the behavior so there is a lack of candidates for a degree 2 polynomial.

The correct interpretation suggests that with the information given none of the equations perfectly aligns with the requirements put forth but choosing m(x) = (x+1)(x−3)(x−5) indicates flawed consideration in the turning aspects.

Therefore, based on the provided options and major analysis, the best option is: m(x) = (x + 1)(x - 3)(x - 5) would be the chosen option even if imperfect in confirming true tuning points.