Solve y = 3x² + 21: When is the Function Positive?

Q: Why doesn't this quadratic have any zeros?

Because $ 3x^2 + 21 = 0 $ would require $ x^2 = -7 $, which has no real solutions. Since squares are never negative, the function never touches the x-axis.

Q: How do I know the parabola opens upward?

Look at the coefficient of $ x^2 $! Since $ a = 3 > 0 $, the parabola opens upward. If a were negative, it would open downward.

Q: How can I visualize this problem?

Imagine a U-shaped curve that starts at point (0, 21) and goes up in both directions. Since the lowest point is at y = 21, the entire curve stays above the x-axis!

Q: Is there a faster way to see this?

Yes! Since $ 3x^2 \geq 0 $ for all x, and we're adding 21, we get $ 3x^2 + 21 \geq 21 > 0 $. The function is always positive!

Quadratic Functions with Positive Analysis

Given the function:

$y=3x^2+21$

Determine for which values of x the following holds:

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

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=3x^2+21$

Determine for which values of x the following holds:

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

Step-by-step solution

To solve the problem, we must determine when the function $f(x) = 3x^2 + 21$ is positive:

Step 1: Analyze the quadratic function $f(x) = ax^2 + bx + c$ . Since $a = 3$ and $b = 0$ , the parabola opens upwards with vertex at $x = 0$ .
Step 2: Compute the function's value at the vertex. For $x = 0$ , $f(0) = 3(0)^2 + 21 = 21$ , which is positive.
Step 3: Understand the behavior for any $x$ . Since $3x^2$ is non-negative and $+21$ added ensures $f(x) > 0$ , the entire graph of $f(x)$ lies above the x-axis.

Therefore, the solution is that for all values of $x$ , the function $f(x) = 3x^2 + 21$ is always greater than 0.

The correct answer is All x.

Final Answer

All x

Key Points to Remember

Essential concepts to master this topic

Opening Direction: When a > 0, parabola opens upward
Minimum Value: Find vertex at x = 0, f(0) = 21
Check: Since minimum is 21 > 0, all values positive ✓

Common Mistakes

Avoid these frequent errors

Setting function equal to zero to find positive values
Don't solve 3x² + 21 = 0 to find when f(x) > 0! This finds zeros, not positive values. The function has no real zeros since 3x² + 21 is always positive. Always analyze the vertex and opening direction instead.

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 doesn't this quadratic have any zeros?

Because $3x^2 + 21 = 0$ would require $x^2 = -7$ , which has no real solutions. Since squares are never negative, the function never touches the x-axis.

How do I know the parabola opens upward?

Look at the coefficient of $x^2$ ! Since $a = 3 > 0$ , the parabola opens upward. If a were negative, it would open downward.

What if the constant term was different?

If the constant was smaller (like +5), the function might still be always positive. If it was negative (like -21), then the function would be negative between the zeros and positive outside them.

How can I visualize this problem?

Imagine a U-shaped curve that starts at point (0, 21) and goes up in both directions. Since the lowest point is at y = 21, the entire curve stays above the x-axis!

Is there a faster way to see this?

Yes! Since $3x^2 \geq 0$ for all x, and we're adding 21, we get $3x^2 + 21 \geq 21 > 0$ . The function is always positive!

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