Theory of Intelligence
11TH•Mathematics•Theory Of Quadratic Functions•Question 2

Step-by-Step Solution

Find the maximum or minimum point by sketching the following quadratic functions. Also find their domain and range:

Step-by-Step Solution

Step 1: Identify Parameters and Vertex

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

a=1,b=2,c=−8.3a = 1, \quad b = 2, \quad c = -8.3

Because a=1>0a = 1 > 0, the parabola opens upwards, so the vertex is an absolute minimum point.

The xx-coordinate of the vertex hh:

h=−b2a=−22(1)=−1h = -\frac{b}{2a} = -\frac{2}{2(1)} = -1

The yy-coordinate kk:

k=f(−1)=(−1)2+2(−1)−8.3=1−2−8.3=−9.3k = f(-1) = (-1)^2 + 2(-1) - 8.3 = 1 - 2 - 8.3 = -9.3

The vertex is located at (−1,−9.3)(-1, -9.3).

Step 2: Intercepts & Axis of Symmetry

  • yy-intercept: f(0)=−8.3  ⟹  (0,−8.3)f(0) = -8.3 \implies (0, -8.3).
  • xx-intercepts: Set f(x)=(x+1)2−9.3=0  ⟹  (x+1)2=9.3  ⟹  x=−1±9.3≈−1±3.05f(x) = (x + 1)^2 - 9.3 = 0 \implies (x + 1)^2 = 9.3 \implies x = -1 \pm \sqrt{9.3} \approx -1 \pm 3.05, giving x≈2.05x \approx 2.05 and x≈−4.05x \approx -4.05.
  • Axis of Symmetry: x=−1x = -1

Step 3: Domain and Range

  • Domain: Defined for all real inputs: Domain=R=(−∞,∞)\text{Domain} = \mathbb{R} = (-\infty, \infty)
  • Range: The lowest point on the curve is y=−9.3y = -9.3, and the graph opens upwards without bound: Range=[−9.3,∞)\text{Range} = [-9.3, \infty)

Discuss theoretical concepts, mathematical formulas, and architectural foundations.

Supports inline code with `backticks`
No comments yet. Be the first to start the discussion!