Determine x Values for y = 2x² + 6 When Positive

Q: Why doesn't this parabola ever touch the x-axis?

Because the vertex is at (0, 6), which is 6 units above the x-axis. Since the parabola opens upward, this is the lowest point, so the function never reaches zero.

Q: What if the constant term was negative instead of +6?

If we had $ y = 2x^2 - 6 $, the vertex would be at (0, -6). The parabola would still open upward but would cross the x-axis at two points, so it wouldn't always be positive.

Q: Do I need to find where the function equals zero first?

No! The question asks where f(x) > 0, not where f(x) = 0. Since this parabola has no real roots and opens upward, it's always positive.

Q: How do I verify that 'all x' is really the answer?

Test several values! Try x = -2: $ y = 2(-2)^2 + 6 = 14 > 0 $. Try x = 5: $ y = 2(5)^2 + 6 = 56 > 0 $. Every value you test will be positive!

Quadratic Functions with Constant Positivity

Given the function:

$y=2x^2+6$

Determine for which values of x is $f\left(x\right) > 0$ true

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Given the function:

$y=2x^2+6$

Determine for which values of x is $f\left(x\right) > 0$ true

Step-by-step solution

To solve this problem, let's analyze the quadratic function $y = 2x^2 + 6$ :

The function $y = 2x^2 + 6$ is a parabola opening upwards because the coefficient of $x^2$ is positive ( $a = 2$ ).
A parabola that opens upwards has its vertex as the lowest point.
Since the function is in the form $y = ax^2 + c$ with no $x$ term, there are no real roots (our function has no $b$ term, hence no rotations impacting x-intercepts).
We evaluate $f(x) > 0$ : Since the constant term $c = 6$ is positive, this defines a vertical shift of the parabola entirely above the x-axis.
This implies $y = 2x^2 + 6$ is always greater than zero for any value of $x$ .

Therefore, the solution to the problem is all x.

Final Answer

All x

Key Points to Remember

Essential concepts to master this topic

Parabola Direction: Positive coefficient of x² means upward opening
Vertex Analysis: Find minimum at x = 0: y = 2(0)² + 6 = 6
Check: Test any value like x = 1: y = 2(1)² + 6 = 8 > 0 ✓

Common Mistakes

Avoid these frequent errors

Setting the function equal to zero to find roots
Don't solve 2x² + 6 = 0 looking for x-intercepts = no real solutions exist! This creates confusion when the question asks where f(x) > 0. Always analyze the vertex and parabola direction first to determine if the function stays 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 doesn't this parabola ever touch the x-axis?

Because the vertex is at (0, 6), which is 6 units above the x-axis. Since the parabola opens upward, this is the lowest point, so the function never reaches zero.

How can I tell if a quadratic is always positive without graphing?

Check two things: 1) The coefficient of x² is positive (parabola opens up), and 2) The vertex y-value is positive. If both are true, the function is always positive!

What if the constant term was negative instead of +6?

If we had $y = 2x^2 - 6$ , the vertex would be at (0, -6). The parabola would still open upward but would cross the x-axis at two points, so it wouldn't always be positive.

Do I need to find where the function equals zero first?

No! The question asks where f(x) > 0, not where f(x) = 0. Since this parabola has no real roots and opens upward, it's always positive.

How do I verify that 'all x' is really the answer?

Test several values! Try x = -2: $y = 2(-2)^2 + 6 = 14 > 0$ . Try x = 5: $y = 2(5)^2 + 6 = 56 > 0$ . Every value you test will be positive!

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