Question

What is the end behavior of the graph of f(x) = x5 – 8x4 + 16x3?
Answer: 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 = ___

Answer 1

Answer:

Step-by-step explanation:

Please write f(x) = x5 – 8x4 + 16x3   as   f(x) = x^5 – 8x^4 + 16x^3.  " ^ " denotes exponentiation.

We really don't care what the function looks like when near x = 0 or small values of x.  This being an odd powered polynomial, we know immediately that as x increases from left to right, the function increases, mostly in Quadrant III, and then continues to increase without bound in Quadrant I.

Your answer choice, B, is correct.  Just a different way of stating the same thing as I have detailed above.

Answer 2

Answer: End behaviour is,

f(x) => -∞ as x => -∞;

f(x) => +∞ as x => +∞

The graph touches, but does not cross, the x–axis at x = 4

The graph of the function crosses the x–axis at x = 0

Step-by-step explanation:

Given function is,

$f(x)=x^5-8x^4+16x^3----(1)$

The degree of f(x) is 5 ( odd ) with positive leading coefficient,

Hence, the end behaviour of f(x) is,

f(x) => -∞ as x => -∞;

f(x) => +∞ as x => +∞

Now, from equation (1),

$f(x)=x^3(x^2-8x+16)$

If f(x) = 0,

$x^3(x^2-8x+16)=0$

$\implies x^3=0\text{ or }x^2-8x+16=0$

$x^3=0\text{ or }(x-4)^2=0$

$x^3=0\text{ or }x-4=0$

$\implies x=0\text{ or }x=4$

Now, the multiplicity of 4 is 2 ( even )

Thus, the graph touches, but does not cross, the x–axis at x = 4

Also, the multiplicity of 0 is 3 ( odd )

Hence, the graph of the function crosses the x–axis at x = 0