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

Q: How do I know which direction the parabola opens?

Look at the coefficient of $ x^2 $! If it's positive (like our a = 2), the parabola opens upward. If negative, it opens downward.

Q: What does the +3 do to the graph?

The +3 shifts the entire parabola up by 3 units. So instead of the vertex being at (0, 0), it's now at (0, 3). Think of it as lifting the whole curve!

Q: How can I tell the difference between similar graphs?

Focus on the vertex location first! For $ y = 2x^2 + 3 $, the vertex must be at (0, 3). Then check that it opens upward since a = 2 > 0.

Q: Should I test points to verify my answer?

Absolutely! Try easy values like x = 1: $ y = 2(1)^2 + 3 = 5 $. The correct graph should pass through (1, 5) and (-1, 5).

Quadratic Functions with Vertical Transformations

Match the function $y=2x^2+3$

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

00:06 Let's match the function to the right graph.

00:09 Because the coefficient of X squared is positive, the graph opens upward. It's like a happy face.

00:16 Next, let's find where the graph crosses the Y-axis.

00:20 Substitute X equals zero to see where it intersects the Y-axis.

00:26 This point is where our graph crosses the Y-axis.

00:29 Based on the function type and the Y-intercept,

00:33 we conclude there are no points where it intersects the X-axis.

00:38 Now, let's draw the graph using the function type and our intersection point.

00:43 And that's how we solve this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Match the function $y=2x^2+3$

to the corresponding graph.

Step-by-step solution

To solve this problem, we need to match the quadratic function $y = 2x^2 + 3$ with one of the graph choices.

First, identify the characteristics of the parabola:

The standard form of the function is $y = 2x^2 + 3$ , which is already in vertex form for vertical shift.
The parabola opens upwards since the coefficient of $x^2$ is positive ( $a = 2$ ).
The vertex of the parabola is at $(0, 3)$ , not subject to any horizontal shifts. The only transformation from $y = x^2$ is the vertical shift by 3 units up.

Now, assess the graph choices:

Look for a parabola that is centered on the vertical axis (origin along the x-axis) and opens upwards.
Among the provided graphs, the one depicting an upright parabola with vertex at $(0, 3)$ should correspond to our function.

The correct choice is graph 3, as it aligns with our function's characteristics: opening upwards, vertex located at $(0, 3)$ .

Therefore, the solution to the problem is graph 3.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = 2x^2 + 3$ has vertex at (0, 3)
Technique: Coefficient a = 2 > 0 means parabola opens upward
Check: Graph passes through (0, 3) and (1, 5) when x = 1 ✓

Common Mistakes

Avoid these frequent errors

Confusing the y-intercept with the vertex location
Don't think the +3 means the vertex is at (3, 0) = wrong graph choice! The constant term shifts the parabola vertically, not horizontally. Always remember that $y = ax^2 + k$ has vertex at (0, k).

Practice Quiz

Test your knowledge with interactive questions

Which chart represents the function $$ y=x^2-9 $$ ?

FAQ

Everything you need to know about this question

How do I know which direction the parabola opens?

Look at the coefficient of $x^2$ ! If it's positive (like our a = 2), the parabola opens upward. If negative, it opens downward.

What does the +3 do to the graph?

The +3 shifts the entire parabola up by 3 units. So instead of the vertex being at (0, 0), it's now at (0, 3). Think of it as lifting the whole curve!

How can I tell the difference between similar graphs?

Focus on the vertex location first! For $y = 2x^2 + 3$ , the vertex must be at (0, 3). Then check that it opens upward since a = 2 > 0.

Should I test points to verify my answer?

Absolutely! Try easy values like x = 1: $y = 2(1)^2 + 3 = 5$ . The correct graph should pass through (1, 5) and (-1, 5).

What if I picked the wrong graph?

No worries! Go back and check: Does your chosen graph have vertex at (0, 3)? Does it open upward? If not, look for the graph that matches both criteria.

🌟 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