Theory of Intelligence
11TH•Mathematics•Theory Of Quadratic Functions•Exercise 3.1

Introduction

Practice set 3.1 for Theory of Quadratic Functions (completing square, sketching, inverse, absolute value).

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Question 1

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

Select Part:Part (i) of 6
(i)

f(x)=x2+6x+13f(x)=x^2+6x+13

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Step-by-Step Solution

Question 1 • Part (i)

verifiedVerified Answer

Find whether the function has a maximum or minimum value by completing the square, find that value, and the value of xx at which it occurs:

f(x)=x2+6x+13f(x) = x^2 + 6x + 13
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Interactive Lab•Interactive

Interactive Graphical Simulator

functionsQuadratic Extrema Laboratory • Exercise 3.1

Q1 (i): f(x) = x² + 6x + 13

Vertex form transformation, axis of symmetry, and global domain & range analysis.

Min = 4 at x = -3
visibilityArrange Display View:
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f(x)=x2+6x+13f(x) = x^2 + 6x + 13•Minimum Vertex Form
Probe: (-3.00, 4.00)
Range: [4, ∞)Axis: x = -3MIN: (-3, 4)
Live Parabola Evaluation:
f(-3.00) = 1·(-3.00)² + (6)·(-3.00) + (13) = 4
adjustMinimum Vertex Point
(-3, 4)
Extrema Value:4
vertical_align_centerAxis of Symmetry
x=−3x = -3
Formula:x=−b2ax = -\frac{b}{2a}
Domain & Range:
Domain:R=(−∞,∞)\mathbb{R} = (-\infty, \infty)
Range:[4,∞)[4, \infty)
★ Continuous across all x
exploreProbe: x = -3.00•f(x) = 4.00•At Vertex!
x = -8x = 2

Step-by-Step Solution

Step 1: Analyze the Leading Coefficient

In the standard quadratic form f(x)=ax2+bx+cf(x) = ax^2 + bx + c:

a=1,b=6,c=13a = 1, \quad b = 6, \quad c = 13

Because the leading coefficient a=1>0a = 1 > 0, the parabola opens upwards (∪\cup). Therefore, the function has an absolute minimum value at its vertex.

Step 2: Complete the Square

Group the xx-terms and add/subtract the square of half the linear coefficient (b2)2=(62)2=32=9\left(\frac{b}{2}\right)^2 = \left(\frac{6}{2}\right)^2 = 3^2 = 9:

f(x)=(x2+6x)+13=(x2+6x+9)+13−9=(x+3)2+4\begin{aligned} f(x) &= (x^2 + 6x) + 13 \\ &= (x^2 + 6x + 9) + 13 - 9 \\ &= (x + 3)^2 + 4 \end{aligned}

Step 3: Determine the Extremum Point

For all real numbers x∈Rx \in \mathbb{R}, a real square is always non-negative:

(x+3)2≥0(x + 3)^2 \ge 0

Equality (x+3)2=0(x + 3)^2 = 0 is attained if and only if:

x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3

Step 4: Calculate the Minimum Value

Substituting x=−3x = -3:

f(−3)=(−3+3)2+4=0+4=4f(-3) = (-3 + 3)^2 + 4 = 0 + 4 = 4

Verification Table

xxCalculation (x+3)2+4(x+3)^2 + 4f(x)f(x)Note
−5-5(−2)2+4=4+4(-2)^2 + 4 = 4 + 488Left branch (>4> 4)
−3-3(0)2+4=0+4(0)^2 + 4 = 0 + 444Absolute Minimum (Vertex)
−1-1(2)2+4=4+4(2)^2 + 4 = 4 + 488Right branch (Symmetric point)
00(3)2+4=9+4(3)^2 + 4 = 9 + 41313yy-intercept

Final Answer

Minimum value=4atx=−3\boxed{\text{Minimum value} = 4 \quad \text{at} \quad x = -3}
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