Question

which statement describes the graph of $f(x) = -4x^3 - 28x^2 - 32x + 64$?

the graph crosses
the $x$-axis at
$x = -4$ and
touches the $x$-axis
at $x = 1$.

the graph touches
the $x$-axis at
$x = -4$ and
crosses the $x$-axis
at $x = 1$.

the graph touches
the $x$-axis at
$x = 4$ and crosses
the $x$-axis at
$x = -1$.

the graph crosses
the $x$-axis at
$x = 4$ and touches
the $x$-axis at
$x = -1$.

Explanation:

Step 1: Factor the polynomial

We start with the function \( f(x) = -4x^3 - 28x^2 - 32x + 64 \). First, factor out a common factor of \(-4\):
\( f(x) = -4(x^3 + 7x^2 + 8x - 16) \)
Now, we try to find a root of the cubic polynomial \( x^3 + 7x^2 + 8x - 16 \) by testing possible rational roots. The possible rational roots are factors of \( 16 \) divided by factors of \( 1 \), so \( \pm1, \pm2, \pm4, \pm8, \pm16 \).
Testing \( x = 1 \): \( 1^3 + 7(1)^2 + 8(1) - 16 = 1 + 7 + 8 - 16 = 0 \). So \( x = 1 \) is a root.
We can then factor the cubic as \( (x - 1)(x^2 + 8x + 16) \) (using polynomial division or synthetic division).
Notice that \( x^2 + 8x + 16 = (x + 4)^2 \). So the factored form of \( f(x) \) is:
\( f(x) = -4(x - 1)(x + 4)^2 \)

Step 2: Analyze the roots

To find where the graph intersects the \( x \)-axis, we set \( f(x) = 0 \):
\( -4(x - 1)(x + 4)^2 = 0 \)
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Answer:

The graph crosses the \( x \)-axis at \( x = -4 \) and touches the \( x \)-axis at \( x = 1 \). (First option)