Solve y = x² + 8x + 20: Finding Values Where Function is Positive

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

This quadratic has a minimum value of 4 at its vertex. Since the parabola opens upward and never goes below y = 4, it never crosses the x-axis (where y = 0).

Q: How do I know the parabola opens upward?

Look at the coefficient of $ x^2 $! When a > 0 (here a = 1), the parabola opens upward like a smile. When a

Q: What if the minimum value was negative?

If the minimum was negative, the function would be negative near the vertex and positive farther away. You'd need to find the zeros to determine exactly where it changes sign.

Q: Can I use the discriminant to solve this?

The discriminant \( b^2 - 4ac = 64 - 80 = -16 confirms no real zeros exist, but vertex analysis gives you the complete picture of where the function is positive.

Q: Is there a faster way than finding the vertex?

For this type of problem, vertex analysis is the most reliable method. It tells you both where the minimum occurs and what that minimum value is.

Quadratic Functions with Vertex Analysis

Look at the following function:

$y=x^2+8x+20$

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 following function:

$y=x^2+8x+20$

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

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

Step-by-step solution

The function given is $y = x^2 + 8x + 20$ . This is a quadratic function where the coefficient of $x^2$ (which is $a = 1$ ) is positive, indicating the parabola opens upwards.

Let’s calculate the vertex to find the minimum value of $y$ . The vertex of a parabola described by $y = ax^2 + bx + c$ is found at $x = -\frac{b}{2a}$ .

Here, $a = 1$ , $b = 8$ . So the vertex is at:

x = -\frac{8}{2 \times 1} = -4

Substitute $x = -4$ into the function $y = x^2 + 8x + 20$ to calculate the minimum value of $y$ .

y = (-4)^2 + 8(-4) + 20 = 16 - 32 + 20 = 4

The minimum value of the function is $y = 4$ at $x = -4$ .

Given the opening direction of the parabola and the positive minimum value, the function $f(x) = x^2 + 8x + 20$ is always greater than 0.

Thus, the function is positive for all values of $x$ .

Final Answer

The function is positive for all values of $x$ .

Key Points to Remember

Essential concepts to master this topic

Rule: When a > 0, parabola opens upward with minimum value
Technique: Find vertex at $x = -\frac{b}{2a} = -\frac{8}{2} = -4$
Check: Minimum value y = 4 > 0, so function is always positive ✓

Common Mistakes

Avoid these frequent errors

Finding zeros instead of analyzing minimum value
Don't try to solve x² + 8x + 20 = 0 to find where function is positive = discriminant approach gives wrong focus! This function has no real zeros since minimum is 4. Always find the vertex first to determine the function's minimum value.

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?

This quadratic has a minimum value of 4 at its vertex. Since the parabola opens upward and never goes below y = 4, it never crosses the x-axis (where y = 0).

How do I know the parabola opens upward?

Look at the coefficient of $x^2$ ! When a > 0 (here a = 1), the parabola opens upward like a smile. When a < 0, it opens downward.

What if the minimum value was negative?

If the minimum was negative, the function would be negative near the vertex and positive farther away. You'd need to find the zeros to determine exactly where it changes sign.

Can I use the discriminant to solve this?

The discriminant $b^2 - 4ac = 64 - 80 = -16 < 0$ confirms no real zeros exist, but vertex analysis gives you the complete picture of where the function is positive.

Is there a faster way than finding the vertex?

For this type of problem, vertex analysis is the most reliable method. It tells you both where the minimum occurs and what that minimum value is.

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Solve y = x² + 8x + 20: Finding Values Where Function is Positive

Quadratic Functions with Vertex Analysis

❤️ 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 doesn't this quadratic have any zeros?

How do I know the parabola opens upward?

What if the minimum value was negative?

Can I use the discriminant to solve this?

Is there a faster way than finding the vertex?

🌟 Unlock Your Math Potential

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