\( 1<\quad \begin{array}{l}\text { Find the range of the quadratic function and the maximum or minimum value of the function. Identify the intervals over } \\ \text { which the function is increasing or decreasing. } \\ y=2 x^{2}+5 x+3\end{array} \)

We are given the quadratic function   y = 2x² + 5x + 3. Step 1. Find the Vertex (the minimum point for this upward-opening parabola). • For a quadratic function y = ax² + bx + c, the x-coordinate of the vertex is given by x = –b/(2a). Here, a = 2 and b = 5, so   x = –5 / (2 × 2) = –5/4. • To find the y-coordinate, substitute x = –5/4 into the function:   y = 2(–5/4)² + 5(–5/4) + 3. Calculate each term:   – (–5/4)²: (–5/4)² = 25/16,   – First term: 2 × (25/16) = 50/16 = 25/8,   – Second term: 5 × (–5/4) = –25/4,   – Third term: 3 (which can be written as 24/8 to match denominators). Combine the terms:   y = 25/8 – (25/4) + 24/8    (Note that 25/4 = 50/8),   y = (25 – 50 + 24) / 8,   y = (–1) / 8,   y = –1/8. Thus, the vertex is at (–5/4, –1/8), which is the minimum point. Step 2. Determine the Range Since the coefficient of x² is positive (2 > 0), the parabola opens upward and has a minimum point at y = –1/8. Therefore, the range of the function is:   Range: [–1/8, ∞). Step 3. Identify the Increasing and Decreasing Intervals • Find the derivative of y with respect to x:   y' = d/dx (2x² + 5x + 3) = 4x + 5. • Set the derivative equal to zero to locate the critical point:   4x + 5 = 0 ⟹ x = –5/4. Examine the sign of y' on either side of x = –5/4:  • For x < –5/4, choose a test point (e.g., x = –2):    y' = 4(–2) + 5 = –8 + 5 = –3  (results in a negative derivative)    This means y is decreasing on (–∞, –5/4).  • For x > –5/4, choose a test point (e.g., x = 0):    y' = 4(0) + 5 = 5       (results in a positive derivative)    This means y is increasing on (–5/4, ∞). Summary: • The function has a minimum value of –1/8 at x = –5/4. • The range of the function is [–1/8, ∞). • The function is decreasing on the interval (–∞, –5/4) and increasing on the interval (–5/4, ∞).