Questions: HW16 Rational Functions: Problem 2 (5 points) Suppose (f(x)=frac2 x^2-4x^3-6). 1. Evaluate (f(5)= (46) /(119)) 2. What are the (x)-intercept of (f(x)) ? Write your answer as an ordered pair. (sqrt(2),0),(-sqrt(2),0) 3. What is the (y)-intercept of (f(x)) ? Write your answer as an ordered pair. ((0,(2) /(3))) 4. Write the equation of the vertical asymptote of (f(x)) : 5. Write the equation of the horizontal asymptote of (f(x)) :

Solution Steps

1. To evaluate $f(5)$, substitute $x = 5$ into the function $f(x) = \frac{2x^2 - 4}{x^3 - 6}$ and simplify the expression.
2. To find the $x$-intercepts, set the numerator of the function equal to zero and solve for $x$. The $x$-intercepts occur where the function equals zero, i.e., where $2x^2 - 4 = 0$.
3. To find the $y$-intercept, evaluate the function at $x = 0$. This is done by substituting $x = 0$ into the function and simplifying.

To evaluate $f(5)$, we substitute $x = 5$ into the function: $$f(5) = \frac{2(5)^2 - 4}{(5)^3 - 6} = \frac{50 - 4}{125 - 6} = \frac{46}{119}$$

The $x$-intercepts occur where the function equals zero, which is when the numerator is zero: $$2x^2 - 4 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2}$$ Thus, the $x$-intercepts are $(-\sqrt{2}, 0)$ and $(\sqrt{2}, 0)$.

To find the $y$-intercept, we evaluate the function at $x = 0$: $$f(0) = \frac{2(0)^2 - 4}{(0)^3 - 6} = \frac{-4}{-6} = \frac{2}{3}$$ Thus, the $y$-intercept is $(0, \frac{2}{3})$.

Final Answer