Match the Quadratic Function y = x²/4 + 2 to Its Graph

Q: How do I find the vertex from $ y = \frac{x^2}{4} + 2 $ ?

This is already in vertex form $ y = a(x-h)^2 + k $! Here, h = 0 and k = 2, so the vertex is at (0, 2).

Q: Why does the coefficient $ \frac{1}{4} $ make the parabola wider?

When 0 , the parabola stretches horizontally, making it wider. Since $ \frac{1}{4} , this parabola is 4 times wider than \( y = x^2 $.

Q: How can I tell which direction the parabola opens?

Look at the coefficient of $ x^2 $! Since $ \frac{1}{4} > 0 $ (positive), the parabola opens upward. Negative coefficients open downward.

Q: What's the difference between this and $ y = x^2 + 2 $ ?

Both have vertex at (0, 2), but $ y = \frac{x^2}{4} + 2 $ is much wider. At x = 4, our function gives y = 6, while $ y = x^2 + 2 $ gives y = 18!

Q: How do I verify I picked the right graph?

Check key points! The vertex should be at (0, 2), and test another point like x = 2: $ y = \frac{4}{4} + 2 = 3 $. The point (2, 3) should be on your chosen graph.

Quadratic Functions with Vertex Form Analysis

One function

$y=\frac{x^2}{4}+2$

to the corresponding 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 Match the function to the appropriate graph

00:03 Notice that X's coefficient is squared and positive, so the function is smiling

00:09 We want to find the intersection point with the Y-axis

00:13 We'll substitute X=0 and solve to find the intersection point with Y-axis

00:19 This is the intersection point with the Y-axis

00:22 According to the function type and intersection point

00:25 We can conclude there's no intersection point with the X-axis

00:28 Let's draw the graph according to the function type and intersection point we found

00:31 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

One function

$y=\frac{x^2}{4}+2$

to the corresponding graph:

Step-by-step solution

The function given is $y = \frac{x^2}{4} + 2$ , which is a quadratic function with a vertex at $(0, 2)$ . The function is in the form $y = a(x-h)^2 + k$ , where $a = \frac{1}{4}$ , $h = 0$ , and $k = 2$ . This tells us that the parabola opens upwards with its vertex at $(0, 2)$ , and it's wider than the standard parabola $y = x^2$ because $\frac{1}{4}$ is less than 1.

To find the correct graph, look for the one featuring a vertex at $(0, 2)$ with an upward opening, and wider spread due to the smaller coefficient. When comparing the graphs, the graph labeled as choice 1 clearly shows these characteristics, indicating the correct match for the function.

Therefore, the solution corresponds to the graph labeled as choice 1.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = a(x-h)^2 + k$ identifies vertex at (h, k)
Technique: Coefficient $a = \frac{1}{4}$ makes parabola wider than $y = x^2$
Check: Test vertex point (0, 2): $y = \frac{0^2}{4} + 2 = 2$ ✓

Common Mistakes

Avoid these frequent errors

Confusing vertex coordinates with y-intercept
Don't think the vertex is at (0, 0) just because there's no x-term = wrong graph selection! The +2 shifts the entire parabola up 2 units. Always identify that the vertex is at (0, 2), not the origin.

Practice Quiz

Test your knowledge with interactive questions

Find the ascending area of the function

$$ f(x)=2x^2 $$

FAQ

Everything you need to know about this question

How do I find the vertex from $y = \frac{x^2}{4} + 2$ ?

This is already in vertex form $y = a(x-h)^2 + k$ ! Here, h = 0 and k = 2, so the vertex is at (0, 2).

Why does the coefficient $\frac{1}{4}$ make the parabola wider?

When 0 < a < 1, the parabola stretches horizontally, making it wider. Since $\frac{1}{4} < 1$ , this parabola is 4 times wider than $y = x^2$ .

How can I tell which direction the parabola opens?

Look at the coefficient of $x^2$ ! Since $\frac{1}{4} > 0$ (positive), the parabola opens upward. Negative coefficients open downward.

What's the difference between this and $y = x^2 + 2$ ?

Both have vertex at (0, 2), but $y = \frac{x^2}{4} + 2$ is much wider. At x = 4, our function gives y = 6, while $y = x^2 + 2$ gives y = 18!

How do I verify I picked the right graph?

Check key points! The vertex should be at (0, 2), and test another point like x = 2: $y = \frac{4}{4} + 2 = 3$ . The point (2, 3) should be on your chosen graph.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Parabola Families 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

Match the Quadratic Function y = x²/4 + 2 to Its Graph

Quadratic Functions with Vertex Form Analysis

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

How do I find the vertex from $y = \frac{x^2}{4} + 2$ ?

Why does the coefficient $\frac{1}{4}$ make the parabola wider?

How can I tell which direction the parabola opens?

What's the difference between this and $y = x^2 + 2$ ?

How do I verify I picked the right graph?

🌟 Unlock Your Math Potential

More Questions

Parabola of the Form y=x²+c

Match the Quadratic Function y = 6x² to its Corresponding Graph

Match the Quadratic Function y = -6x² to its Corresponding Graph

Match the Quadratic Function y = 2x² + 3 to Its Correct Graph

Match the Quadratic Function y = x² + 9 to Its Correct Graph

Match the Quadratic Function y = x²/4 + 2 to Its Graph

Quadratic Functions with Vertex Form Analysis

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

How do I find the vertex from y=x24+2 y = \frac{x^2}{4} + 2 y=4x2​+2?

Why does the coefficient 14 \frac{1}{4} 41​ make the parabola wider?

How can I tell which direction the parabola opens?

What's the difference between this and y=x2+2 y = x^2 + 2 y=x2+2?

How do I verify I picked the right graph?

🌟 Unlock Your Math Potential

More Questions

Parabola of the Form y=x²+c

Match the Quadratic Function y = 6x² to its Corresponding Graph

Match the Quadratic Function y = -6x² to its Corresponding Graph

Match the Quadratic Function y = 2x² + 3 to Its Correct Graph

Match the Quadratic Function y = x² + 9 to Its Correct Graph

How do I find the vertex from $y = \frac{x^2}{4} + 2$ ?

Why does the coefficient $\frac{1}{4}$ make the parabola wider?

What's the difference between this and $y = x^2 + 2$ ?