What is the minimum value of `2x^(2) -3x + 5` ?

Options

A. 0
B. `3//4`
C. `31//4`
D. `31//8`

Answer

To find the minimum value of the quadratic function \( f(x) = 2x^2 - 3x + 5 \), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic function is in the form \( ax^2 + bx + c \), where:
- \( a = 2 \)
- \( b = -3 \)
- \( c = 5 \)

### Step 2: Check if the parabola opens upwards
Since \( a = 2 \) is greater than 0, the parabola opens upwards, indicating that there is a minimum value.

### Step 3: Find the x-coordinate of the vertex
The x-coordinate of the vertex (which gives the minimum value for a parabola that opens upwards) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( b \) and \( a \):
\[ x = -\frac{-3}{2 \times 2} = \frac{3}{4} \]

### Step 4: Calculate the minimum value
Now, we substitute \( x = \frac{3}{4} \) back into the function to find the minimum value:
\[ f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 5 \]
Calculating \( \left(\frac{3}{4}\right)^2 \):
\[ \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]
Now substituting back:
\[ f\left(\frac{3}{4}\right) = 2 \times \frac{9}{16} - 3 \times \frac{3}{4} + 5 \]
Calculating each term:
\[ = \frac{18}{16} - \frac{9}{4} + 5 \]
Converting \( \frac{9}{4} \) to have a common denominator of 16:
\[ \frac{9}{4} = \frac{36}{16} \]
And converting 5:
\[ 5 = \frac{80}{16} \]
Now substituting these values:
\[ f\left(\frac{3}{4}\right) = \frac{18}{16} - \frac{36}{16} + \frac{80}{16} \]
Combining the fractions:
\[ = \frac{18 - 36 + 80}{16} = \frac{62}{16} = \frac{31}{8} \]

### Conclusion
The minimum value of the function \( 2x^2 - 3x + 5 \) is \( \frac{31}{8} \).

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