Domain Analysis of y=-1/6x²-5/7: Finding Valid Input Values

Q: What does 'positive domain' and 'negative domain' mean exactly?

The positive domain includes all x-values where y is positive (above x-axis). The negative domain includes all x-values where y is negative (below x-axis).

Q: How do I know if the parabola opens up or down?

Look at the coefficient of $ x^2 $! If it's positive, the parabola opens upward. If it's negative (like $ -\frac{1}{6} $), it opens downward.

Q: Why is the vertex at x = 0 in this problem?

Since there's no x-term in $ y = -\frac{1}{6}x^2 - \frac{5}{7} $, the parabola is symmetric about the y-axis. The vertex formula gives x = 0 when the middle coefficient is zero.

Q: What if the vertex y-value was positive instead?

If the vertex had a positive y-value and the parabola opened downward, there would be some x-values giving positive y (near the vertex) and others giving negative y (far from vertex).

Q: How can I double-check my answer?

Test a few x-values! Try x = 1: $ y = -\frac{1}{6}(1)^2 - \frac{5}{7} = -\frac{1}{6} - \frac{5}{7} $ which is negative. Try x = -2: similar negative result. This confirms all y-values are negative.

Quadratic Functions with Domain Classification

Find the positive and negative domains of the function below:

$y=-\frac{1}{6}x^2-\frac{5}{7}$

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the positive and negative domains of the function below:

$y=-\frac{1}{6}x^2-\frac{5}{7}$

Step-by-step solution

To solve the problem, we proceed as follows:

The quadratic function is given as $y = -\frac{1}{6}x^2 - \frac{5}{7}$ .
The coefficient $a = -\frac{1}{6}$ is negative, indicating the parabola opens downward.
The function has no $x$ -term ( $b = 0$ ), so the vertex $x$ -coordinate is $0$ .
Evaluating $y$ at $x = 0$ , we have: $y = -\frac{5}{7}$ , which is negative.
Since the parabola opens downward and $y$ at the vertex is negative, $y$ remains negative for all $x$ values.
Thus, there is no positive domain.
For $x < 0$ , $y$ is negative for all $x$ .

In conclusion, the function remains negative for all $x < 0$ and is also negative for all else. The answer choice matches:

$x < 0 :$ all $x$

$x > 0 :$ none

Final Answer

$x < 0 :$ all $x$

$x > 0 :$ none

Key Points to Remember

Essential concepts to master this topic

Parabola Direction: Negative coefficient means downward opening parabola
Vertex Analysis: At x = 0, y = -5/7 (maximum value)
Domain Check: Since maximum is negative, all y-values are negative ✓

Common Mistakes

Avoid these frequent errors

Confusing positive/negative domain with x-values instead of y-values
Don't look at where x is positive or negative = wrong interpretation! The question asks about domains where the function output y is positive or negative. Always analyze the y-values (function outputs) to determine positive and negative domains.

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

What does 'positive domain' and 'negative domain' mean exactly?

The positive domain includes all x-values where y is positive (above x-axis). The negative domain includes all x-values where y is negative (below x-axis).

How do I know if the parabola opens up or down?

Look at the coefficient of $x^2$ ! If it's positive, the parabola opens upward. If it's negative (like $-\frac{1}{6}$ ), it opens downward.

Why is the vertex at x = 0 in this problem?

Since there's no x-term in $y = -\frac{1}{6}x^2 - \frac{5}{7}$ , the parabola is symmetric about the y-axis. The vertex formula gives x = 0 when the middle coefficient is zero.

What if the vertex y-value was positive instead?

If the vertex had a positive y-value and the parabola opened downward, there would be some x-values giving positive y (near the vertex) and others giving negative y (far from vertex).

How can I double-check my answer?

Test a few x-values! Try x = 1: $y = -\frac{1}{6}(1)^2 - \frac{5}{7} = -\frac{1}{6} - \frac{5}{7}$ which is negative. Try x = -2: similar negative result. This confirms all y-values are negative.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 The Quadratic Function questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime