Decipher the Graph's Story: Which Equation Represents This Function?

Q: How can I tell if a parabola opens up or down just from the equation?

Look at the coefficient of $ x^2 $! If it's positive (like in $ y = x^2 $ or $ y = 3x^2 $), the parabola opens upward. If it's negative (like in $ y = -x^2 $), it opens downward.

Q: Why does the graph look like an upside-down U?

The negative sign in $ y = -x^2 $ flips the normal parabola! Instead of going up as x gets farther from zero, the y-values go down, creating that upside-down U shape.

Q: How do I know this isn't y = 3x² or y = ½x²?

Both $ y = 3x^2 $ and $ y = \frac{1}{2}x^2 $ have positive coefficients, so they open upward like a regular U. Our graph opens downward, so it must have a negative coefficient.

Q: What makes this y = -x² instead of y = -2x² or y = -½x²?

The graph passes through specific points! At $ x = 1 $, $ y = -1 $. Let's check: $ y = -(1)^2 = -1 $ ✓. Other equations like $ y = -2x^2 $ would give $ y = -2 $ at $ x = 1 $.

Q: Can I test points to verify my answer?

Absolutely! Pick any point from the graph and substitute into your chosen equation. For example, at $ x = 2 $, the graph shows $ y = -4 $. Testing $ y = -x^2 $: $ -(2)^2 = -4 $ ✓

Quadratic Functions with Downward-Opening Parabolas

Which of the following equations corresponds to the function represented in the graph?

❤️ Continue Your Math Journey!

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

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the appropriate equation for the function in the graph

00:03 Let's find points on the graph

00:32 Let's observe the pattern, and deduce the function's equation from it

00:38 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Which of the following equations corresponds to the function represented in the graph?

Step-by-step solution

The given graph is a parabola that opens downwards and is symmetrical with respect to the y-axis, with its vertex at the origin. This is characteristic of a standard quadratic function of the form $y = -ax^2$ , where $a$ is positive, indicating the parabola opens downwards because $a$ is negative in the form $y = -x^2$ .

Let's compare the graph with the choices given:

The first choice $y = x^2$ opens upwards, so it doesn't match.
The second choice $y = -x^2$ opens downwards, matching the graph.
The third choice $y = 3x^2$ opens upwards and is more compressed, so it doesn’t match.
The fourth choice $y = \frac{1}{2}x^2$ opens upwards and is wider, so it doesn’t match.

Therefore, the equation that corresponds to the graph is $y = -x^2$ .

Final Answer

$y=-x^2$

Key Points to Remember

Essential concepts to master this topic

Direction Rule: Negative coefficient makes parabola open downward
Recognition: When $a < 0$ in $y = ax^2$ , parabola opens down
Check: Test point (1, -1): $-1 = -(1)^2 = -1$ ✓

Common Mistakes

Avoid these frequent errors

Confusing upward and downward opening directions
Don't assume all parabolas open upward like $y = x^2$ = wrong graph match! The negative sign in front of $x^2$ flips the parabola upside down. Always check the coefficient sign: positive opens up, negative opens down.

Practice Quiz

Test your knowledge with interactive questions

Determine whether the following table represents a constant function:

FAQ

Everything you need to know about this question

How can I tell if a parabola opens up or down just from the equation?

Look at the coefficient of $x^2$ ! If it's positive (like in $y = x^2$ or $y = 3x^2$ ), the parabola opens upward. If it's negative (like in $y = -x^2$ ), it opens downward.

Why does the graph look like an upside-down U?

The negative sign in $y = -x^2$ flips the normal parabola! Instead of going up as x gets farther from zero, the y-values go down, creating that upside-down U shape.

How do I know this isn't y = 3x² or y = ½x²?

Both $y = 3x^2$ and $y = \frac{1}{2}x^2$ have positive coefficients, so they open upward like a regular U. Our graph opens downward, so it must have a negative coefficient.

What makes this y = -x² instead of y = -2x² or y = -½x²?

The graph passes through specific points! At $x = 1$ , $y = -1$ . Let's check: $y = -(1)^2 = -1$ ✓. Other equations like $y = -2x^2$ would give $y = -2$ at $x = 1$ .

Can I test points to verify my answer?

Absolutely! Pick any point from the graph and substitute into your chosen equation. For example, at $x = 2$ , the graph shows $y = -4$ . Testing $y = -x^2$ : $-(2)^2 = -4$ ✓

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Functions 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