David graphs a polynomial function with the features listed below:
- Zeros at [tex]x=-1, x=2, x=3[/tex]
- [tex]y[/tex]-intercept at [tex](0, -6)[/tex]
- End behavior of [tex]f(x) \rightarrow \infty[/tex] as [tex]x \rightarrow -\infty[/tex] and [tex]f(x) \rightarrow -\infty[/tex] as [tex]x \rightarrow \infty[/tex]

Which function did David graph?
A. [tex]f(x)=-x^3-4x^2-x+6[/tex]
B. [tex]f(x)=x^3-4x^2+x+6[/tex]
C. [tex]f(x)=-x^3+4x^2-x-6[/tex]
D. [tex]f(x)=x^3+4x^2+x-6[/tex]



Answer :

To figure out which function David graphed, let's follow these steps:

1. Identify the polynomial function structure:
- The zeros of the polynomial are given: [tex]\( x = -1 \)[/tex], [tex]\( x = 2 \)[/tex], and [tex]\( x = 3 \)[/tex].
- Therefore, the polynomial can be expressed as [tex]\( f(x) = k (x + 1)(x - 2)(x - 3) \)[/tex], where [tex]\( k \)[/tex] is a constant.

2. Determine the constant [tex]\( k \)[/tex] using the [tex]\( y \)[/tex]-intercept:
- Given the [tex]\( y \)[/tex]-intercept at [tex]\( (0, -6) \)[/tex], we substitute [tex]\( x = 0 \)[/tex] and [tex]\( f(0) = -6 \)[/tex] into the polynomial.
- So, [tex]\( -6 = k (0 + 1)(0 - 2)(0 - 3) \)[/tex].
- Simplifying, [tex]\( -6 = k (1)(-2)(-3) \)[/tex].
- Since [tex]\( 1 \times -2 \times -3 = 6 \)[/tex], we have [tex]\( -6 = k \times 6 \)[/tex].
- Solving for [tex]\( k \)[/tex], [tex]\( k = -1 \)[/tex].

3. Formulate the polynomial equation:
- Substituting [tex]\( k = -1 \)[/tex] into the polynomial, we get [tex]\( f(x) = -1 (x + 1)(x - 2)(x - 3) \)[/tex].

4. Expand the polynomial:
- Start by expanding [tex]\( (x + 1)(x - 2)(x - 3) \)[/tex].
- First, calculate [tex]\( (x - 2)(x - 3) \)[/tex]:
[tex]\[ (x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6. \][/tex]
- Next, multiply the result by [tex]\( (x + 1) \)[/tex]:
[tex]\[ (x + 1)(x^2 - 5x + 6) = x(x^2 - 5x + 6) + 1(x^2 - 5x + 6) = x^3 - 5x^2 + 6x + x^2 - 5x + 6. \][/tex]
- Combining like terms:
[tex]\[ x^3 - 4x^2 + x + 6. \][/tex]
- Since [tex]\( k = -1 \)[/tex], we multiply the entire polynomial by [tex]\(-1\)[/tex]:
[tex]\[ f(x) = -1 \times (x^3 - 4x^2 + x + 6) = -x^3 + 4x^2 - x - 6. \][/tex]

5. Check the end behavior:
- For the polynomial [tex]\( f(x) = -x^3 + 4x^2 - x - 6 \)[/tex]:
- As [tex]\( x \to -\infty \)[/tex], [tex]\( -x^3 \to \infty \)[/tex], so [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( -x^3 \to -\infty \)[/tex], so [tex]\( f(x) \to -\infty \)[/tex].

This matches the given end behavior.

Therefore, David graphed the function [tex]\( f(x) = -x^3 + 4x^2 - x - 6 \)[/tex], which corresponds to option C.

So, the correct answer is:
C. [tex]\( f(x) = -x^3 + 4x^2 - x - 6 \)[/tex].