minimum value of \( 2 x^{2}+3 x+5 \)

1. Recognize that the quadratic function is given by \[ f(x) = 2x^2 + 3x + 5. \] 2. The quadratic function \( f(x) = ax^2 + bx + c \) reaches its minimum (or maximum) at \[ x = -\frac{b}{2a}. \] Here, \( a = 2 \) and \( b = 3 \), so \[ x = -\frac{3}{2 \cdot 2} = -\frac{3}{4}. \] 3. Substitute \( x = -\frac{3}{4} \) into the original function to find the minimum value: \[ f\left(-\frac{3}{4}\right) = 2\left(-\frac{3}{4}\right)^2 + 3\left(-\frac{3}{4}\right) + 5. \] 4. Calculate each term: - First term: \[ 2\left(-\frac{3}{4}\right)^2 = 2\left(\frac{9}{16}\right) = \frac{18}{16} = \frac{9}{8}. \] - Second term: \[ 3\left(-\frac{3}{4}\right) = -\frac{9}{4}. \] - Third term: \[ 5 = \frac{40}{8} \quad \text{(written with a common denominator for ease)}. \] 5. Combine the terms using a common denominator: \[ f\left(-\frac{3}{4}\right) = \frac{9}{8} - \frac{18}{8} + \frac{40}{8}. \] 6. Add the numerators: \[ \frac{9 - 18 + 40}{8} = \frac{31}{8}. \] 7. Thus, the minimum value of the function is \[ \frac{31}{8}, \] which occurs at \( x = -\frac{3}{4} \).