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Which could be the function graphed below? On a coordinate plane, a curve opens down and to the right. In quadrants111and444, the curve starts on theyyy-axis in quadrant444and goes through thexxx-axis into quadrant111.

Question

Which could be the function graphed below?

On a coordinate plane, a curve opens down and to the right. In quadrants 1 and 4, the curve starts on the y-axis in quadrant 4 and goes through the x-axis into quadrant 1.

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Solution

To determine which function could be represented by the described graph, let's analyze the characteristics of the curve based on the given description.

Step 1: Define the Problem

Identify the key components of the question:

  • The curve opens down and to the right.
  • It starts on the y-axis in quadrant IV and goes through the x-axis into quadrant I.

Step 2: Break Down the Problem

  • Curve Direction: The curve opens downwards, indicating a negative leading coefficient if it's a polynomial.
  • Quadrant Movement: The curve starts in quadrant IV and moves to quadrant I, suggesting it crosses the x-axis from negative to positive y-values.

Step 3: Apply Relevant Concepts

  • Function Type: The description suggests a function that decreases as it moves from quadrant IV to quadrant I. This could be a linear function with a negative slope or a part of a quadratic function with a negative leading coefficient.
  • Crossing the Axes: The function crosses the x-axis, indicating it has a real root.

Step 4: Analysis, Verify and Summarize

  • Linear Function: A linear function like y=mx+c y = -mx + c (where m>0 m > 0 ) could fit if it crosses the x-axis and has a negative slope.
  • Quadratic Function: A quadratic function like y=ax2+bx+c y = -ax^2 + bx + c (where a>0 a > 0 ) could also fit if it opens downwards and crosses the x-axis.

Final Answer

The function could be a linear function with a negative slope or a quadratic function with a negative leading coefficient, such as:

  • Linear: y=mx+c y = -mx + c (where m>0 m > 0 )
  • Quadratic: y=ax2+bx+c y = -ax^2 + bx + c (where a>0 a > 0 )

This problem has been solved

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