What is the end behavior of the graph of f(x) = x5 – 8x4 + 16x3?
b.) f(x) => -∞ as x => -∞; f(x) => +∞ as x => +∞
the graph touches, but does not cross, the x–axis at x =__
the graph of the function crosses the x–axis at x =

f(x) = x⁵ – 8x⁴ + 16x³

Does not cross the x-axis at x = 4.

Crosses the x-axis at x = 0

This is about the end behavior of a graph of a function at the end of the x-axis.

We are given the function;

f(x) = x⁵ - 8x⁴ + 16x³

f(x) = x⁵ - 8x⁴ + 16x³

Let's factorize it first to get;

f(x) = x³(x² – 8x + 16)

(x² – 8x + 16) is a perfect square trinomial and can be expressed as (x – 4)(x - 4).

Thus;

f(x) = x³ (x – 4)(x - 4)

f(x) = x³ (x – 4)(x - 4)

The roots are at f(x) = 0;

The roots are; x³ = 0; (x – 4) = 0 ; (x - 4) = 0

This means the roots of f(x) are; x=0 and x=4.  x = 4

This means the graph has a repeated root and so it will touch the x-axis but not at the repeated root of x=4.

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Step-by-step explanation:

f(x) = x⁵ – 8x⁴ + 16x³

As x approaches +∞, the highest term, x⁵, approaches +∞.

As x approaches -∞, x⁵ approaches -∞ (a negative number raised to an odd exponent is also negative).

Now let's factor:

f(x) = x³ (x² – 8x + 16)

f(x) = x³ (x – 4)²

f(x) has roots at x=0 and x=4.  x=4 is a repeated root (because it's squared), so the graph touches the x-axis but does not cross at x=4.

The graph crosses the x-axis at x=0.

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