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1) Write a polynomial equation for a function with a graph that bounces off the x-axis at (-1,0),

crosses it at (4,0), and goes through the point (-2, -18).

User Migmaker
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1 Answer

2 votes

Answer:

f(x) = 3x³ -6x² -21x -12

Explanation:

You want a polynomial function that touches the x-axis at x = -1, crosses it at x = 4, and goes through the point (-2, -18).

Factored form

Each zero (p) of a polynomial corresponds to a linear factor (x -p). If the zero has even multiplicity, the function will not change sign there, but will touch the x-axis and "bounce".

The requirements indicate a factor (x +1) with even multiplicity, and a factor (x -4). The least-degree such polynomial will have the factored form ...

f(x) = a(x +1)²(x -4)

Scaling

The value of this polynomial at x = -2 is ...

f(-2) = a(-2 +1)²(-2 -4) = -6a

We want this to be -18, so ...

-6a = -18 ⇒ a = 3

Then the factored polynomial is ...

f(x) = 3(x +1)²(x -4)

When we multiply this out, we get ...

f(x) = 3x³ -6x² -21x -12

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Additional comment

The attached graph verifies the desired characteristics.

1) Write a polynomial equation for a function with a graph that bounces off the x-example-1
User Jack Murphy
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7.9k points

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