Converting a Semicircular Graph to Its Algebraic Function Representation

Q: How can I tell if it's $ y = x^2 $ or $ y = -x^2 $ ?

Look at the opening direction! If the semicircle opens upward (like a smile), it's $ y = x^2 $. If it opened downward (like a frown), it would be $ y = -x^2 $.

Q: Why isn't this a full parabola?

The graph only shows the upper half where y ≥ 0. For $ y = x^2 $, all y-values are non-negative, so we only see the semicircle portion above the x-axis.

Q: How do I know the vertex is at the origin?

The graph appears centered on the coordinate axes. For $ y = x^2 $, the vertex is always at (0,0) when there are no horizontal or vertical shifts.

Q: Could this be a different degree polynomial?

No! The smooth curved shape and symmetric nature clearly indicate a quadratic function. Linear functions are straight lines, and cubic functions have different curvature patterns.

Q: What if the semicircle was shifted up or down?

Then it would be $ y = x^2 + c $ where c is the vertical shift. Since this graph passes through the origin, there's no vertical shift, so c = 0.

Quadratic Functions with Semicircle Graphs

Find the corresponding algebraic representation for the function

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Choose the appropriate algebraic representation for the function

00:04 In a smiling function, the coefficient of X squared is positive

00:08 Conversely, in a sad function, the coefficient is negative

00:12 Our function is smiling, so a negative coefficient is not possible

00:19 In this case, X is not squared, so the function will be linear

00:31 When the function is linear, if you input a negative number, it stays negative

00:34 Our function is always positive, so this cannot be

00:43 The same case applies in this option

00:56 Let's check this option with a negative number

01:00 In this option, the function is always positive

01:04 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

Find the corresponding algebraic representation for the function

Step-by-step solution

In this problem, we are tasked with identifying the algebraic representation of a function given a graphical depiction. Given the problem's indication that we are dealing with parabolas, particularly those of the form $y = x^2 + c$ , we need to examine the provided graph for features typical of this family of functions.

The graph structure in the problem suggests a parabolic curve, centered symmetrically, which is indicative of the simplest unmodified parabola, $y = x^2$ . The vertex likely lies at the origin, and the parabola opens upwards, a key characteristic of the function $y = x^2$ when the coefficient of $x^2$ is positive and equal to 1.

Upon reviewing the multiple-choice options, the expression that corresponds to this graph is:

Option 1: $y = x^2$

Therefore, the algebraic representation that corresponds to the function is $y = x^2$ .

Final Answer

$y=x^2$

Key Points to Remember

Essential concepts to master this topic

Shape Recognition: Semicircle opening upward indicates positive quadratic function
Vertex Analysis: Graph centered at origin means $y = x^2$ with no shifts
Verification: Test points like (1,1) and (-1,1) confirm symmetry about y-axis ✓

Common Mistakes

Avoid these frequent errors

Confusing semicircle orientation with function sign
Don't assume a semicircle opening upward means $y = -x^2$ = downward parabola! The graph shows the upper half of a circle, which corresponds to the positive values of $y = x^2$ . Always match the visual direction with the algebraic sign.

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 can I tell if it's $y = x^2$ or $y = -x^2$ ?

Look at the opening direction! If the semicircle opens upward (like a smile), it's $y = x^2$ . If it opened downward (like a frown), it would be $y = -x^2$ .

Why isn't this a full parabola?

The graph only shows the upper half where y ≥ 0. For $y = x^2$ , all y-values are non-negative, so we only see the semicircle portion above the x-axis.

How do I know the vertex is at the origin?

The graph appears centered on the coordinate axes. For $y = x^2$ , the vertex is always at (0,0) when there are no horizontal or vertical shifts.

Could this be a different degree polynomial?

No! The smooth curved shape and symmetric nature clearly indicate a quadratic function. Linear functions are straight lines, and cubic functions have different curvature patterns.

What if the semicircle was shifted up or down?

Then it would be $y = x^2 + c$ where c is the vertical shift. Since this graph passes through the origin, there's no vertical shift, so c = 0.

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Converting a Semicircular Graph to Its Algebraic Function Representation

Quadratic Functions with Semicircle Graphs

❤️ 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 can I tell if it's $y = x^2$ or $y = -x^2$ ?

Why isn't this a full parabola?

How do I know the vertex is at the origin?

Could this be a different degree polynomial?

What if the semicircle was shifted up or down?

🌟 Unlock Your Math Potential

More Questions

Parabola of the Form y=x²+c

Find the Algebraic Equation: Parabola Passing Through (4,0)

Quadratic Function: Converting Graph to Algebraic Form at Point (-2)

Converting a Semicircular Graph to Its Algebraic Function Representation

Quadratic Functions with Semicircle Graphs

❤️ 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 can I tell if it's y=x2 y = x^2 y=x2 or y=−x2 y = -x^2 y=−x2?

Why isn't this a full parabola?

How do I know the vertex is at the origin?

Could this be a different degree polynomial?

What if the semicircle was shifted up or down?

🌟 Unlock Your Math Potential

More Questions

Parabola of the Form y=x²+c

Find the Algebraic Equation: Parabola Passing Through (4,0)

Quadratic Function: Converting Graph to Algebraic Form at Point (-2)

How can I tell if it's $y = x^2$ or $y = -x^2$ ?