Theory of Intelligence
11TH•Mathematics•Theory Of Quadratic Functions•Question 1

Step-by-Step Solution

Find the maximum or minimum value of the following quadratic functions by completing square:

Step-by-Step Solution

Step 1: Analyze the Leading Coefficient

Comparing with f(x)=ax2+bx+cf(x) = ax^2 + bx + c:

a=−1,b=−3,c=−5a = -1, \quad b = -3, \quad c = -5

Because a=−1<0a = -1 < 0, the parabola opens downwards (∩\cap). Therefore, the function achieves an absolute maximum value at its vertex.

Step 2: Complete the Square

Factor out −1-1 from the xx-terms:

f(x)=−(x2+3x)−5f(x) = -(x^2 + 3x) - 5

Complete the square inside the parenthesis by adding (32)2=94\left(\frac{3}{2}\right)^2 = \frac{9}{4}. Balancing outside by adding 94\frac{9}{4}:

f(x)=−(x2+3x+94)−5+94=−(x+32)2−204+94=−(x+32)2−114\begin{aligned} f(x) &= -\left(x^2 + 3x + \frac{9}{4}\right) - 5 + \frac{9}{4} \\ &= -\left(x + \frac{3}{2}\right)^2 - \frac{20}{4} + \frac{9}{4} \\ &= -\left(x + \frac{3}{2}\right)^2 - \frac{11}{4} \end{aligned}

Step 3: Determine the Extremum Point

Because −(x+32)2≤0-\left(x + \frac{3}{2}\right)^2 \le 0 for all real xx, the maximum is achieved when the squared expression vanishes:

x+32=0  ⟹  x=−32=−1.5x + \frac{3}{2} = 0 \implies x = -\frac{3}{2} = -1.5

Step 4: Calculate the Maximum Value

Substituting x=−32x = -\frac{3}{2}:

f(−32)=−0−114=−114=−2.75f\left(-\frac{3}{2}\right) = -0 - \frac{11}{4} = -\frac{11}{4} = -2.75

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