Question

graph the function, considering the domain, critical points, symmetry, regions where the function is increasing or decreasing, inflection points, regions where the function is concave upward or concave downward, intercepts where possible, and asymptotes where applicable.

$f(x)=\frac{-2x + 7}{2x + 5}$

choose the graph of the function.

○ a.

○ b.

○ c.

○ d.

Explanation:

Step1: Find domain

Set denominator ≠ 0: $2x + 5 ≠ 0 ⇒ x ≠ -\frac{5}{2}$. Domain: $\mathbb{R} \setminus \{-\frac{5}{2}\}$.

Step2: Find asymptotes

Vertical asymptote: $x = -\frac{5}{2}$.
Horizontal asymptote: $\lim_{x→±∞}\frac{-2x+7}{2x+5} = \frac{-2}{2} = -1 ⇒ y = -1$.

Step3: Find intercepts

x-intercept: Set $f(x)=0 ⇒ -2x+7=0 ⇒ x=\frac{7}{2}=3.5$.
y-intercept: Set $x=0 ⇒ f(0)=\frac{7}{5}=1.4$.

Step4: Check monotonicity

First derivative: $f'(x)=\frac{(-2)(2x+5)-(-2x+7)(2)}{(2x+5)^2}=\frac{-4x-10+4x-14}{(2x+5)^2}=\frac{-24}{(2x+5)^2} < 0$ for all $x≠-\frac{5}{2}$. So function is decreasing everywhere in domain.

Step5: Check concavity

Second derivative: $f''(x)=\frac{(-24)(-2)(2)}{(2x+5)^3}=\frac{96}{(2x+5)^3}$.

• When $x > -\frac{5}{2}$, $f''(x) > 0$ (concave up).
• When $x < -\frac{5}{2}$, $f''(x) < 0$ (concave down).

Answer:

(Note: Since the graphs are not fully visible, based on the properties: decreasing everywhere, vertical asymptote at $x=-2.5$, horizontal asymptote at $y=-1$, x-intercept at 3.5, y-intercept at 1.4, concave up right of $x=-2.5$ and concave down left of $x=-2.5$, the correct graph should match these features. Assuming the standard options for this function, the answer is typically the graph with these characteristics.)

(If forced to select from A-D based on common textbook problems for this function, the answer is usually the graph that shows the right branch decreasing from above $y=-1$ to below, left branch decreasing from below $y=-1$ to above, with intercepts at (3.5,0) and (0,1.4).)