Explore the Quadratic: Find Positive and Negative Domains of y = -1/2x² + x - 1

Find the positive and negative domains of the following function:

y=12x2+x1 y=-\frac{1}{2}x^2+x-1

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Step-by-step written solution

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1

Understand the problem

Find the positive and negative domains of the following function:

y=12x2+x1 y=-\frac{1}{2}x^2+x-1

2

Step-by-step solution

To solve this problem, we'll determine where the quadratic function y=12x2+x1 y = -\frac{1}{2}x^2 + x - 1 is positive and negative.

Step 1: Find the roots of the quadratic equation.
We'll use the quadratic formula, x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=12 a = -\frac{1}{2} , b=1 b = 1 , and c=1 c = -1 .

Calculate the discriminant b24ac=124(12)(1)=12=1 b^2 - 4ac = 1^2 - 4(-\frac{1}{2})(-1) = 1 - 2 = -1 .
Notice that the discriminant is negative, meaning the quadratic equation has no real roots.

Step 2: Determine the sign of the quadratic function.
Since there are no real roots, the quadratic does not intersect the x-axis. Since a=12 a = -\frac{1}{2} , which is negative, the parabola opens downwards. Without real roots, it means it is always negative for all values of x x .

Conclusion: The function is negative for all x x .

Therefore, the positive and negative domains of the function are:

x<0: x < 0 : for all x x

x>0: x > 0 : none

3

Final Answer

x<0: x < 0 : for all x x

x>0: x > 0 : none

Practice Quiz

Test your knowledge with interactive questions

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 \( f\left(x\right) > 0 \).

AAABBBCCCX

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