Question

If \(k\) is the minimum value of the algebraic expression \(x^2 + 3x + 2\) attained at \(x = a\), then \((k,a)=\)

• A. \(\left(\frac12,\frac34\right)\)
• B. \(\left(\frac14,\frac32\right)\)
• C. \(\left(\frac18,\frac{3}{16}\right)\)
• D. \(\left(-\frac14,-\frac32\right)\)

Hint:

For any quadratic \(ax^2+bx+c\), the vertex gives minimum/maximum: \[ x=\frac{-b}{2a}, \quad y_{\min/\max}=f(x). \]

Answer 1

The Correct Option is D

Concept: For a quadratic expression \[ ax^2+bx+c \] the minimum value occurs at \[ x=\frac{-b}{2a} \]

Step 1: Find the value of \(a\). \[ x^2+3x+2 \] Here \(a=1, b=3\). So the value of \(x\) at minimum is: \[ x=\frac{-3}{2} \] Thus, \[ a=-\frac{3}{2} \]

Step 2: Find the minimum value \(k\). \[ k=\left(-\frac{3}{2}\right)^2+3\left(-\frac{3}{2}\right)+2 \] \[ =\frac{9}{4}-\frac{9}{2}+2 \] Convert to common denominator: \[ =\frac{9}{4}-\frac{18}{4}+\frac{8}{4} \] \[ =\frac{-1}{4} \] So, \[ k=-\frac14 \] \[ \boxed{\left(k,a\right)=\left(-\frac14,-\frac32\right)} \]