Question

Which of the following could represent the graph of f(x) = x4 + x3 – 8x2 – 12x?

Answer 1

Answer:

The graph of f(x) is given by the 2nd graph.

                             

Step-by-step explanation:

Here, the given function is,

f(x) = $x^{4} +x^{3} -8x^{2} - 12x$

so $\dfrac {df(x)}{dx}$

= f'(x) = $4x^{3} + 3x^{2} - 16x -12$------------(1)

        = $4x(x^{2} - 4) + 3(x^{2} - 4)$

        = $(4x + 3)(x + 2)(x - 2)$

so f'(x) = 0 for x = $\frac{-3}{4}$ , -2, 2

so, f"(x) = $\dfrac {d^{2}(f(x))}{dx^{2}}$--------(from (1))

            = $12x^{2} +6x - 16$ -------------(2)

So, from (2) for x = -2

f"(x) = $12 \times ({-2})^{2} +6 \times ({-2}) -16$

      = 20 > 0

So, f(x) has a minima at x = -2 .

So, among the given graphs only the middle-most or the 2nd graph satisfies this criteria. So, the graph of f(x) is given by the 2nd graph.

                             

Answer 2

Answer:

Second graph.

Step-by-step explanation:

An way to find the correct graph is finding all four roots of the equation when, $f(x)=0$, using Ruffini's Rule, which consists in arranging all coefficients to find all the zeros through division.

First, we can extract a common factor:

$x^{4}+x^{3}-8x^{2}-12x=0\\x(x^{3}+x^{2}-8x-12)=0$

Applying the null factor rule we already have the first solution:

$x_{1}=0\\x^{3}+x^{2}-8x-12=0$

Now, we apply Ruffini's Rule to solve the third grade equation by taking coefficients only, the image attached shows the process. Basically we have to find one divisor of 12 that gives zero as a result of the division.

Now, we evaluate the function at $x=1$, if gives us $y=-2$, the second graph would be the solution:

Applying Ruffini's rules we find that the solution of the third-grade equation are:

$x_{2}=-2\\ x_{3}=-3\\x_{4}=-2$

So, now, we look the graph that has all these intercerption points with x-axis.

That gives us graph three and four as possible, then we can graph the function to ensure the right answer.

Now, using the graphing method, we see that second option is correct, because it has the same inflection patterns.