Solve the Quadratic Inequality: When Is y = -5x² - 10 Less Than Zero?

Q: How do I know if a parabola opens up or down?

Look at the coefficient of $ x^2 $! If it's positive, the parabola opens upward (U-shape). If it's negative like our -5, it opens downward (∩-shape).

Q: Why is the vertex at x = 0 for this problem?

The vertex formula is $ x = -\frac{b}{2a} $. In $ y = -5x^2 - 10 $, we have b = 0, so $ x = -\frac{0}{2(-5)} = 0 $.

Q: What if the parabola touched the x-axis instead?

If the vertex was at $ y = 0 $, then \( f(x) would be true for no values (since a downward parabola touching y = 0 would be ≤ 0, not

Q: Can I solve this by setting the function equal to zero?

That would find where $ f(x) = 0 $, not where $ f(x) ! For \( -5x^2 - 10 = 0 $, you'd get $ x^2 = -2 $, which has no real solutions - confirming the parabola never touches the x-axis.

Q: How do I check that ALL values of x work?

Since the maximum value is -10 (at the vertex), and the parabola only goes downward from there, every point must be negative. Test a few: $ f(1) = -15 $, $ f(-2) = -30 $ - all negative!

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Look at the function below:

$y=-5x^2-10$

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

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

2

Step-by-step solution

Let's solve this step-by-step.

First, identify the key features of the parabola represented by $y = -5x^2 - 10$ . It is a downward-opening parabola because the coefficient of $x^2$ , which is -5, is negative.
Since the parabola opens downwards, the vertex provides the highest point on the graph. The general formula for the vertex $x$ -coordinate of the quadratic function $ax^2 + bx + c$ is $-\frac{b}{2a}$ . However, as mentioned, $b = 0$ , so the vertex is at $x = 0$ .
Substitute $x = 0$ back into the function to find the $y$ -coordinate of the vertex:
$y = -5(0)^2 - 10 = -10$ .
This implies that the entire parabola is situated below the y-axis since the maximum value (vertex) is -10, which is already less than zero.
Thus, the function $f(x) = -5x^2 - 10$ is always less than zero for all $x$ . This is because the vertex, the highest point, is also negative, and it only opens downward from there.

Therefore, the solution to the problem is that $f(x) < 0$ for all values of $x$ .

3

Final Answer

All values of $x$

Key Points to Remember

Essential concepts to master this topic

Shape Rule: Negative coefficient means parabola opens downward always
Vertex Method: Find maximum at $x = 0$ , giving $y = -10$
Check: Test any x-value: $-5(1)^2 - 10 = -15 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Thinking the parabola crosses the x-axis
Don't assume every parabola crosses y = 0 just because it's quadratic = missing that this one stays entirely below zero! The vertex is at y = -10, and since it opens downward, all values are negative. Always find the vertex first to see the parabola's highest or lowest point.

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

How do I know if a parabola opens up or down?

+

Look at the coefficient of $x^2$ ! If it's positive, the parabola opens upward (U-shape). If it's negative like our -5, it opens downward (∩-shape).

Why is the vertex at x = 0 for this problem?

+

The vertex formula is $x = -\frac{b}{2a}$ . In $y = -5x^2 - 10$ , we have b = 0, so $x = -\frac{0}{2(-5)} = 0$ .

What if the parabola touched the x-axis instead?

+

If the vertex was at $y = 0$ , then $f(x) < 0$ would be true for no values (since a downward parabola touching y = 0 would be ≤ 0, not < 0).

Can I solve this by setting the function equal to zero?

+

That would find where $f(x) = 0$ , not where $f(x) < 0$ ! For $-5x^2 - 10 = 0$ , you'd get $x^2 = -2$ , which has no real solutions - confirming the parabola never touches the x-axis.

How do I check that ALL values of x work?

+

Since the maximum value is -10 (at the vertex), and the parabola only goes downward from there, every point must be negative. Test a few: $f(1) = -15$ , $f(-2) = -30$ - all negative!

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Solve the Quadratic Inequality: When Is y = -5x² - 10 Less Than Zero?

Quadratic Inequalities with Negative Parabolas

❤️ 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

How do I know if a parabola opens up or down?

Why is the vertex at x = 0 for this problem?

What if the parabola touched the x-axis instead?

Can I solve this by setting the function equal to zero?

How do I check that ALL values of x work?

🌟 Unlock Your Math Potential

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