Determine x: Solving y = -2x² + 8x - 10 for Positive Values

Q: Why does this function never become positive?

Since the parabola opens downward (a = -2 all other points must be even lower (more negative).

Q: How do I know which direction the parabola opens?

Look at the coefficient of $ x^2 $! If it's positive, the parabola opens upward (∪). If it's negative, it opens downward (∩). Here, a = -2

Q: What if I got confused and thought it was positive somewhere?

A common mistake! Remember: for downward parabolas, if the maximum point (vertex) is below the x-axis, the entire function stays below the x-axis. No part of it can be positive.

Q: How do I find the vertex quickly?

Use $ x_v = -\frac{b}{2a} $. Here: $ x_v = -\frac{8}{2(-2)} = 2 $. Then substitute x = 2 back into the original function to get y = -2.

Q: Could this function ever equal zero?

Let's check! Set $ -2x^2 + 8x - 10 = 0 $ and use the discriminant: \( b^2 - 4ac = 64 - 80 = -16 . Since it's negative, there are no real roots, so the function never equals zero either.

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Look at the following function:

$y=-2x^2+8x-10$

Determine for which values of $x$ the following is true:

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

2

Step-by-step solution

To solve this problem, we'll check where the quadratic function $y = -2x^2 + 8x - 10$ is greater than zero.

The steps to solve are as follows:

Step 1: Identify the direction of opening and calculate the vertex.
For the quadratic function $y = ax^2 + bx + c$ with $a = -2$ , $b = 8$ , $c = -10$ , since $a = -2 < 0$ , the parabola opens downwards.
Step 2: Find the vertex, which is the peak of the parabola.
The x-coordinate of the vertex $x_v$ is given by the formula $x_v = -\frac{b}{2a} = -\frac{8}{2 \times (-2)} = 2$ .
Substitute $x = 2$ into the function to find the y-coordinate: $y = -2(2)^2 + 8(2) - 10 = -8 + 16 - 10 = -2$ .
The vertex is at $(2, -2)$ . This indicates that the maximum value of the function is $-2$ , which is less than zero.
Step 3: Analyze the function's values.
Since the maximum point $(2, -2)$ is less than zero and the parabola opens downwards, the function values are always less than or equal to the vertex's y-value, which is $-2$ .
Step 4: Conclusion.
It is clear that the function is never positive at any point in its domain for real values of $x$ .

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

Direction: When a < 0, parabola opens downward with maximum at vertex
Vertex Formula: x-coordinate = -b/2a = -8/(-4) = 2, then find y-value
Check Maximum: At vertex (2, -2), if maximum < 0, then f(x) ≤ 0 everywhere ✓

Common Mistakes

Avoid these frequent errors

Assuming quadratic functions are always positive somewhere
Don't assume every parabola crosses the x-axis = missing negative-only functions! When a downward-opening parabola has its maximum below the x-axis, it's never positive. Always check if the vertex y-value is above or below zero first.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the $$ x $$ -axis.

The parabola's vertex is marked A.

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

FAQ

Everything you need to know about this question

Why does this function never become positive?

+

Since the parabola opens downward (a = -2 < 0), its highest point is the vertex at (2, -2). If the maximum value is -2, which is negative, then all other points must be even lower (more negative).

How do I know which direction the parabola opens?

+

Look at the coefficient of $x^2$ ! If it's positive, the parabola opens upward (∪). If it's negative, it opens downward (∩). Here, a = -2 < 0, so it opens downward.

What if I got confused and thought it was positive somewhere?

+

A common mistake! Remember: for downward parabolas, if the maximum point (vertex) is below the x-axis, the entire function stays below the x-axis. No part of it can be positive.

How do I find the vertex quickly?

+

Use $x_v = -\frac{b}{2a}$ . Here: $x_v = -\frac{8}{2(-2)} = 2$ . Then substitute x = 2 back into the original function to get y = -2.

Could this function ever equal zero?

+

Let's check! Set $-2x^2 + 8x - 10 = 0$ and use the discriminant: $b^2 - 4ac = 64 - 80 = -16 < 0$ . Since it's negative, there are no real roots, so the function never equals zero either.

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Determine x: Solving y = -2x² + 8x - 10 for Positive Values

Quadratic Functions with Downward Opening Analysis

❤️ Continue Your Math Journey!

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why does this function never become positive?

How do I know which direction the parabola opens?

What if I got confused and thought it was positive somewhere?

How do I find the vertex quickly?

Could this function ever equal zero?

🌟 Unlock Your Math Potential

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