Solve y=-3x²+6x-9: Finding x-Values When Function is Positive

Quadratic Functions with Negative Discriminants

Look at the following function:

y=3x2+6x9 y=-3x^2+6x-9

Determine for which x x values the following is true:

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

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Look at the following function:

y=3x2+6x9 y=-3x^2+6x-9

Determine for which x x values the following is true:

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

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify the given quadratic function and its form.
  • Step 2: Determine the nature of the parabola based on the leading coefficient.
  • Step 3: Calculate the discriminant to find possible roots.
  • Step 4: Solve for the roots using the quadratic formula if applicable.
  • Step 5: Analyze the intervals determined by the roots to assess where the function is positive.
  • Step 6: Conclude whether the function is positive over any interval based on the parabola's direction.

Now, let's work through each step:

Step 1: The given function is y=3x2+6x9 y = -3x^2 + 6x - 9 , which is a quadratic in the standard form ax2+bx+c ax^2 + bx + c , where a=3 a = -3 , b=6 b = 6 , and c=9 c = -9 .

Step 2: The coefficient of x2 x^2 is negative (a=3 a = -3 ), indicating the parabola opens downward.

Step 3: The discriminant Δ \Delta for the quadratic equation ax2+bx+c=0 ax^2 + bx + c = 0 is given by Δ=b24ac \Delta = b^2 - 4ac . Calculating this:

Δ=(6)24(3)(9)=36108=72 \Delta = (6)^2 - 4(-3)(-9) = 36 - 108 = -72 .

Step 4: A negative discriminant (Δ<0 \Delta < 0 ) shows that the quadratic equation 3x2+6x9=0 -3x^2 + 6x - 9 = 0 has no real roots. This means the parabola does not intersect the x-axis.

Step 5: Knowing the downward opening parabola and lack of real roots, the parabola lies entirely below the x-axis, and it never becomes positive anywhere.

Step 6: Since the function is always non-positive, we conclude that the function has no positive domain.

Therefore, the solution to the problem is The function has no positive domain.

3

Final Answer

The function has no positive domain.

Key Points to Remember

Essential concepts to master this topic
  • Rule: When a<0 a < 0 , parabola opens downward from vertex
  • Technique: Calculate discriminant: Δ=36108=72 \Delta = 36 - 108 = -72 (no real roots)
  • Check: Negative discriminant + downward parabola = always below x-axis ✓

Common Mistakes

Avoid these frequent errors
  • Assuming quadratic functions always have positive regions
    Don't think every parabola must cross the x-axis and have positive values = wrong analysis! When the discriminant is negative and the parabola opens downward, it never goes above zero. Always check both the direction (coefficient of x²) and discriminant together.

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

FAQ

Everything you need to know about this question

What does it mean when the discriminant is negative?

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A negative discriminant means the parabola never touches the x-axis. There are no real solutions to the equation when f(x)=0 f(x) = 0 , so the function is either always positive or always negative.

How do I know if the function is always positive or always negative?

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Look at the coefficient of x²! If it's positive, the parabola opens upward (always positive with negative discriminant). If it's negative like a=3 a = -3 , it opens downward (always negative).

Can I still find the vertex even with no real roots?

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Yes! The vertex is at x=b2a=62(3)=1 x = -\frac{b}{2a} = -\frac{6}{2(-3)} = 1 . At x=1 x = 1 , y=6 y = -6 , so the vertex is (1, -6).

Why check the discriminant instead of just graphing?

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The discriminant gives you immediate information about roots without graphing. Δ<0 \Delta < 0 means no x-intercepts, Δ=0 \Delta = 0 means one x-intercept, and Δ>0 \Delta > 0 means two x-intercepts.

What if the question asked when f(x) < 0 instead?

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Since this function is always negative, the answer would be 'for all real values of x' or xR x \in \mathbb{R} . The function never equals zero, so it's negative everywhere.

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