Given the function [tex]\( f(x) = -4x^6 + 6x^2 - 52 \)[/tex], choose the correct statement:

A. [tex]\( f(x) \)[/tex] is an even function so both ends of the graph go in opposite directions.
B. The leading coefficient is positive so the left end of the graph goes down.
C. The leading coefficient is positive so the left end of the graph goes up.
D. [tex]\( f(x) \)[/tex] is an even function so both ends of the graph go in the same direction.



Answer :

Alright, let's analyze the function [tex]\( f(x) = -4x^6 + 6x^2 - 52 \)[/tex] step-by-step to address the given options.

### Step 1: Check if the Function is Even
A function [tex]\( f(x) \)[/tex] is even if and only if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex] in its domain. Let's check this:

[tex]\[ f(-x) = -4(-x)^6 + 6(-x)^2 - 52 \][/tex]

Simplify the terms:

[tex]\[ f(-x) = -4x^6 + 6x^2 - 52 \][/tex]

Since:

[tex]\[ f(-x) = f(x) \][/tex]

Hence, [tex]\( f(x) \)[/tex] is an even function.

### Step 2: Determine the Leading Coefficient
The leading term of the polynomial [tex]\( f(x) = -4x^6 + 6x^2 - 52 \)[/tex] is [tex]\( -4x^6 \)[/tex]. The leading coefficient is the coefficient of the term with the highest power. In this case, it is:

[tex]\[ -4 \][/tex]

So, the leading coefficient is negative.

### Step 3: Analyze the Behavior of the Graph
For polynomials, the behavior of the graph as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex] is determined by the leading term. Here the leading term is [tex]\( -4x^6 \)[/tex], which has an even power (6) and a negative coefficient.

- When the degree is even and the leading coefficient is negative, the ends of the graph both go downward.

### Conclusion
Let's evaluate the provided options:

- Option A: [tex]\( f(x) \)[/tex] is an even function, but both ends of the graph do not go in opposite directions. They both go down. Hence, this is incorrect.

- Option B: The leading coefficient is not positive; it is negative. Hence, this is incorrect.

- Option C: The leading coefficient is negative, not positive, and the left end of the graph does not go up. Hence, this is incorrect.

- Option D: [tex]\( f(x) \)[/tex] is an even function, and since the leading coefficient is negative and the degree is even, both ends of the graph go in the same direction (down). Hence, this is correct.

Thus, the correct answer is:

[tex]\[ \boxed{\text{D. } f(x) \text{ is an even function so both ends of the graph go in the same direction.}} \][/tex]