Solve y=-(4x+32)²: Finding Values Where Function is Positive

Q: Why can't this function ever be positive?

The expression $ (4x + 32)^2 $ is always non-negative (zero or positive). When you multiply by -1, you flip the sign, making it always non-positive (zero or negative). There's no way to get a positive result!

Q: What happens at x = -8?

At x = -8, we get $ 4(-8) + 32 = 0 $, so $ y = -(0)^2 = 0 $. This is the maximum value of the function, but zero is still not positive.

Q: How do I know this parabola opens downward?

The negative sign in front of the squared term tells you! When the coefficient of the squared term is negative, the parabola opens downward, creating a maximum point instead of a minimum.

Q: Could there be any values where f(x) > 0?

No, never! Since $ (4x + 32)^2 \geq 0 $ for all real x, multiplying by -1 gives us $ -(4x + 32)^2 \leq 0 $. The function can equal zero but never exceed it.

Q: What if I graphed this function?

You'd see a downward-opening parabola with its vertex at (-8, 0). The entire graph would be on or below the x-axis, never above it where y would be positive.

Quadratic Functions with Negative Leading Coefficient

Look at the function below:

$y=-\left(4x+32\right)^2$

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

$f(x) > 0$

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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:

$y=-\left(4x+32\right)^2$

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

$f(x) > 0$

Step-by-step solution

To determine for which values of $x$ the function $y = -\left(4x + 32\right)^2$ is positive, we must analyze the behavior of this function.

The function $y = -\left(4x + 32\right)^2$ is a quadratic function with a negative leading coefficient (the negative sign outside the squared term). This indicates that the parabola opens downwards. Let's break down the expression:

The expression inside the parentheses is $4x + 32$ .
Squaring any real number $\left( 4x + 32 \right)^2$ always results in a non-negative value (it is zero or positive).
Multiplying this non-negative result by $-1$ makes $y$ non-positive (either zero or negative).

Since the smallest value $\left( 4x + 32 \right)^2$ can be is zero (when $x = -8$ ), and all other values will just make it negative when multiplied by $-1$ , $y$ can never be greater than zero for any real number $x$ .

Thus, there are no values of $x$ for which $f(x) > 0$ . Consequently, the answer is:

True for no values of $x$ .

Final Answer

True for no values of $x$

Key Points to Remember

Essential concepts to master this topic

Rule: Negative squared expressions can never be positive
Technique: When $(4x + 32)^2 \geq 0$ , then $-(4x + 32)^2 \leq 0$
Check: Test vertex at x = -8: y = -(0)² = 0, not positive ✓

Common Mistakes

Avoid these frequent errors

Thinking the function is positive when the squared term equals zero
Don't assume y > 0 when (4x + 32)² = 0 because -(0)² = 0, not positive! Zero is neither positive nor negative. Always remember that multiplying any non-negative value by -1 makes it non-positive.

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 can't this function ever be positive?

The expression $(4x + 32)^2$ is always non-negative (zero or positive). When you multiply by -1, you flip the sign, making it always non-positive (zero or negative). There's no way to get a positive result!

What happens at x = -8?

At x = -8, we get $4(-8) + 32 = 0$ , so $y = -(0)^2 = 0$ . This is the maximum value of the function, but zero is still not positive.

How do I know this parabola opens downward?

The negative sign in front of the squared term tells you! When the coefficient of the squared term is negative, the parabola opens downward, creating a maximum point instead of a minimum.

Could there be any values where f(x) > 0?

No, never! Since $(4x + 32)^2 \geq 0$ for all real x, multiplying by -1 gives us $-(4x + 32)^2 \leq 0$ . The function can equal zero but never exceed it.

What if I graphed this function?

You'd see a downward-opening parabola with its vertex at (-8, 0). The entire graph would be on or below the x-axis, never above it where y would be positive.

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Solve y=-(4x+32)²: Finding Values Where Function is Positive

Quadratic Functions with Negative Leading Coefficient

❤️ 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 can't this function ever be positive?

What happens at x = -8?

How do I know this parabola opens downward?

Could there be any values where f(x) > 0?

What if I graphed this function?

🌟 Unlock Your Math Potential

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