Match the Function y = -1/2x² + 4 to Its Corresponding Graph

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

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

Q: What's the easiest way to find the vertex?

For $ y = ax^2 + bx + c $, the vertex x-coordinate is $ x = -\frac{b}{2a} $. Since our equation has no bx term, $ b = 0 $, so the vertex is at $ x = 0 $.

Q: How do I find the y-coordinate of the vertex?

Once you know the x-coordinate of the vertex, substitute it back into the original equation. Here: $ y = -\frac{1}{2}(0)^2 + 4 = 4 $, so the vertex is (0,4).

Q: Why is the constant term important?

The constant term $ c = 4 $ tells you the y-intercept - where the parabola crosses the y-axis. It's also the y-coordinate of the vertex when the parabola is centered at $ x = 0 $.

Q: What if I can't see the graphs clearly?

Focus on the key features: vertex location and opening direction. Check if the highest point is at (0,4) and if the parabola curves downward from there. These two features uniquely identify the correct graph.

Quadratic Functions with Downward Opening

One function

$y=-\frac{1}{2}x^2+4$

to the corresponding graph:

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:08 Let's match the function to the correct graph.

00:11 Notice the coefficient of X squared is negative. So, the graph curves downward.

00:18 Next, we'll find where the graph crosses the Y-axis.

00:22 Set X to zero, and solve to get the Y-axis intersection.

00:28 This point is where the graph meets the Y-axis.

00:34 Let's draw the graph using this point and the shape of the function.

00:39 And that's how we solve the problem!

Step-by-step written solution

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Understand the problem

One function

$y=-\frac{1}{2}x^2+4$

to the corresponding graph:

Step-by-step solution

To solve for the graph that matches the function $y = -\frac{1}{2}x^2 + 4$ , let's analyze the function:

The function $y = -\frac{1}{2}x^2 + 4$ is a parabola in standard form $y = ax^2 + bx + c$ , with $a = -\frac{1}{2}$ , $b = 0$ , and $c = 4$ .
Because $a = -\frac{1}{2}$ is negative, the parabola opens downwards.
The vertex of the parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$ . Here, $b = 0$ , so $x = 0$ .
Substituting $x = 0$ back into the equation gives the vertex's y-coordinate: $y = -\frac{1}{2}(0)^2 + 4 = 4$ .
Thus, the vertex is $(0, 4)$ .

Now, let's match this to the graphs:

We are looking for a graph with a vertex at $(0, 4)$ that opens downwards.
Upon reviewing the graphs in the problem, graph number 1 presents a downward opening parabola with a vertex at the point $(0, 4)$ .

Therefore, the graph that corresponds to $y = -\frac{1}{2}x^2 + 4$ is graph 1.

Thus, the solution to the problem is 1.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Vertex Form: Use $x = -\frac{b}{2a}$ to find vertex x-coordinate
Direction Rule: Negative coefficient $a = -\frac{1}{2}$ means parabola opens downward
Verification: Check vertex (0,4) and downward direction match the graph ✓

Common Mistakes

Avoid these frequent errors

Confusing parabola direction based on coefficient sign
Don't think positive coefficient opens downward = upward opening parabola! When $a = -\frac{1}{2}$ is negative, it opens downward, not upward. Always remember: negative coefficient = downward opening, positive coefficient = upward opening.

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, the parabola opens upward (like a smile). If it's negative like $-\frac{1}{2}$ , it opens downward (like a frown).

What's the easiest way to find the vertex?

For $y = ax^2 + bx + c$ , the vertex x-coordinate is $x = -\frac{b}{2a}$ . Since our equation has no bx term, $b = 0$ , so the vertex is at $x = 0$ .

How do I find the y-coordinate of the vertex?

Once you know the x-coordinate of the vertex, substitute it back into the original equation. Here: $y = -\frac{1}{2}(0)^2 + 4 = 4$ , so the vertex is (0,4).

Why is the constant term important?

The constant term $c = 4$ tells you the y-intercept - where the parabola crosses the y-axis. It's also the y-coordinate of the vertex when the parabola is centered at $x = 0$ .

What if I can't see the graphs clearly?

Focus on the key features: vertex location and opening direction. Check if the highest point is at (0,4) and if the parabola curves downward from there. These two features uniquely identify the correct graph.

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