Graph the function \( f(x)=\frac{x+4}{2 x^{2}-5 x-3} \). What does the shape of the graph betweer the vertical asymptotes resemble? an upside-down letter \( U \) the letter \( U \) the graph of \( y=x^{3} \) the graph of \( y=-x^{3} \)

We begin with the function   f(x) = (x + 4) / (2x² – 5x – 3). Step 1. Factor the Denominator The denominator factors nicely:   2x² – 5x – 3 = 2(x – 3)(x + ½). So the function can be written as   f(x) = (x + 4) / [2(x – 3)(x + ½)]. Step 2. Identify the Vertical Asymptotes The factors in the denominator indicate vertical asymptotes where the function is undefined:   x – 3 = 0 ⟹ x = 3,   x + ½ = 0 ⟹ x = –½. Step 3. Analyze the Behavior Between the Vertical Asymptotes We are interested in the behavior of f(x) on the interval (–½, 3). Note the following: • As x approaches –½ from the right:  • (x + ½) → a very small positive number.  • (x – 3) is negative (since –½ – 3 = –3.5).  • The numerator (x + 4) is positive (for x near –½, x + 4 > 3.5). Thus, f(x) → negative infinity near x = –½⁺. • As x approaches 3 from the left:  • (x – 3) → a very small negative number.  • (x + ½) is positive (since 3 + ½ = 3.5).  • The numerator (x + 4) is positive (close to 7). Thus, f(x) → negative infinity near x = 3⁻. Since f(x) is continuous between –½ and 3 and tends to negative infinity at both endpoints, the function must rise from –∞ to a maximum at some point in between and then fall back to –∞. Step 4. Interpreting the Graph’s Shape This behavior creates a single “hump” in the interval, where the graph has a maximum point and then drops down steeply on both sides (approaching –∞ near the vertical asymptotes). Such a shape, with both ends descending toward –∞ and a peak in the middle, resembles an upside-down letter U. Thus, the graph of f(x) between the vertical asymptotes has the shape of an upside-down U. Answer: an upside-down letter U.