Domain Analysis of y = (1/3)x² + x + 2: Finding Valid Input Values

Find the positive and negative domains of the function:

y=13x2+x+2 y=\frac{1}{3}x^2+x+2

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1

Understand the problem

Find the positive and negative domains of the function:

y=13x2+x+2 y=\frac{1}{3}x^2+x+2

2

Step-by-step solution

To determine the positive and negative domains of the quadratic function y=13x2+x+2 y = \frac{1}{3}x^2 + x + 2 , we will follow these general steps:

  • Find the roots of the equation 13x2+x+2=0 \frac{1}{3}x^2 + x + 2 = 0 .
  • Analyze the sign of the quadratic between and beyond these roots.

First, let's identify the roots of the quadratic function:

Using the quadratic formula x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=13 a = \frac{1}{3} , b=1 b = 1 , and c=2 c = 2 , the discriminant Δ=b24ac=124132=183=53 \Delta = b^2 - 4ac = 1^2 - 4 \cdot \frac{1}{3} \cdot 2 = 1 - \frac{8}{3} = \frac{-5}{3} .

Since the discriminant is negative, the quadratic equation 13x2+x+2=0 \frac{1}{3}x^2 + x + 2 = 0 has no real roots.

Given that the coefficient of x2 x^2 (i.e., 13\frac{1}{3}) is positive, the parabola opens upwards, meaning the entire function is greater than zero for all real values of x x .

Therefore, the positive domain is all real numbers, and there is no negative domain.

Therefore, the solution is: x>0 x > 0 : for all x x ; x<0 x < 0 : none.

3

Final Answer

x>0: x > 0 : for all x

x<0: x < 0 : none

Practice Quiz

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The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of \( x \) where \( f\left(x\right) > 0 \).

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