Solve y=-5x²-10: Finding Where Function is Positive

Q: Why does the parabola never go above the x-axis?

The vertex is at $ (0, -10) $, which is 10 units below the x-axis. Since \( a = -5 , the parabola opens downward, so the vertex is the highest point. If the highest point is negative, the entire function stays negative!

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

Look at the coefficient of $ x^2 $! When $ a (like our \( a = -5 $), the parabola opens downward like an upside-down U. When $ a > 0 $, it opens upward like a regular U.

Q: What if the question asked for f(x) < 0 instead?

Then the answer would be all real numbers! Since our function $ y = -5x^2 - 10 $ is always negative, every x-value makes the function less than zero.

Q: Could this function ever equal zero exactly?

Let's check: $ -5x^2 - 10 = 0 $ means $ -5x^2 = 10 $, so $ x^2 = -2 $. Since we can't take the square root of a negative number in real numbers, this function never equals zero!

Q: How can I visualize this without graphing?

Think of it this way: start at the vertex $ (0, -10) $ which is below the x-axis. Since the parabola opens downward, moving left or right from the vertex takes you even further down, never up toward the x-axis!

Quadratic Functions with Negative Maximum Values

Look at the function below:

$y=-5x^2-10$

Then determine for which values of $x$ the following is true:

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

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the function below:

$y=-5x^2-10$

Then determine for which values of $x$ the following is true:

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

Step-by-step solution

To solve this problem, we need to analyze the quadratic function given by:

1. Step 1: Find the vertex of the parabola.
The formula for a quadratic function is $y = ax^2 + bx + c$ . For our function, $a = -5$ , $b = 0$ , $c = -10$ .
Since $b = 0$ , the x-coordinate of the vertex is $x = 0$ .

2. Step 2: Calculate the y-coordinate of the vertex.
Substitute $x = 0$ into the function:
$y = -5(0)^2 - 10 = -10$ .

3. Step 3: Analyze the parabola.
The vertex is at (0, -10). Since the vertex itself is below the x-axis and the parabola opens downwards (given by $a < 0$ ), the entire parabola is below the x-axis.

As a result, there are no values of $x$ for which the function $y = -5x^2 - 10$ is greater than 0.

Therefore, the correct answer to the problem is that there are no values of $x$ .

Final Answer

No values of $x$

Key Points to Remember

Essential concepts to master this topic

Rule: Downward parabola with negative vertex equals always negative function
Technique: Find vertex at x = 0: y = -5(0)² - 10 = -10
Check: Test any x-value: -5(1)² - 10 = -15 < 0 confirms no positive values ✓

Common Mistakes

Avoid these frequent errors

Setting function equal to zero instead of greater than zero
Don't solve -5x² - 10 = 0 when asked for f(x) > 0! This finds x-intercepts, not where function is positive. Always analyze the parabola's position relative to the x-axis using the vertex and direction of opening.

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 the parabola never go above the x-axis?

The vertex is at $(0, -10)$ , which is 10 units below the x-axis. Since $a = -5 < 0$ , the parabola opens downward, so the vertex is the highest point. If the highest point is negative, the entire function stays negative!

How do I know which direction the parabola opens?

Look at the coefficient of $x^2$ ! When $a < 0$ (like our $a = -5$ ), the parabola opens downward like an upside-down U. When $a > 0$ , it opens upward like a regular U.

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

Then the answer would be all real numbers! Since our function $y = -5x^2 - 10$ is always negative, every x-value makes the function less than zero.

Could this function ever equal zero exactly?

Let's check: $-5x^2 - 10 = 0$ means $-5x^2 = 10$ , so $x^2 = -2$ . Since we can't take the square root of a negative number in real numbers, this function never equals zero!

How can I visualize this without graphing?

Think of it this way: start at the vertex $(0, -10)$ which is below the x-axis. Since the parabola opens downward, moving left or right from the vertex takes you even further down, never up toward the x-axis!

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