Convert Quadratic Graph to Algebraic Function: Point (-6) Analysis

Q: How do I tell if it's +6 or -6 in the equation?

Look at where the parabola crosses the y-axis! If it crosses at y = -6, then your equation is $ y = x^2 - 6 $ because when x = 0, you get y = 0 - 6 = -6.

Q: What's the difference between vertex and y-intercept?

For $ y = x^2 - 6 $, the vertex is at (0, -6) and the y-intercept is also at (0, -6). This only happens when the parabola opens upward with no horizontal shift!

Q: How do I know this isn't y = -6x²?

The graph shows a parabola opening upward, not stretched vertically. If it were $ y = -6x^2 $, it would open downward and be much steeper.

Q: Can I check my answer with another point?

Yes! Try x = 1: $ y = 1^2 - 6 = -5 $. The point (1, -5) should be on your parabola. You can also try x = -1 for the same result.

Q: Why isn't the answer y = x²/6?

That would create a horizontal stretch, making the parabola wider. The graph shows the same width as $ y = x^2 $, just moved down 6 units.

Quadratic Functions with Vertical Transformations

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

00:03 Find the intersection point of the function with the Y-axis

00:10 This is the point

00:14 We'll substitute X=0 in each function and compare the intersection points

00:25 According to this representation, the intersection point is different, so not this representation

00:28 Continue with this method for each representation and check the intersection points

00:39 According to this representation, the intersection point is equal, so this representation is correct

00:50 According to this representation, the intersection point is different, so not this representation

01:00 According to this representation, the intersection point is different, so not this representation

01:07 And this is the solution to the question

Step-by-step written solution

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Understand the problem

Find the corresponding algebraic representation for the function

Step-by-step solution

To solve this problem, we will determine the vertical shift given to the parent function $y = x^2$ to form the observed parabola.

Identify that the problem involves a vertical translation of the parabola $y = x^2$ .
The function takes the form $y = x^2 + c$ , where $c$ indicates the vertical shift.
From the graph given, it is seen that the vertex of the parabola is situated at $y = -6$ when viewed from the intersection with the y-axis.
This downward shift corresponds to the constant $c$ being negative, specifically $c = -6$ .
By this observation, the function becomes $y = x^2 - 6$ .

Therefore, the solution to the problem is $y = x^2 - 6$ , matching choice 2.

Final Answer

$y=x^2-6$

Key Points to Remember

Essential concepts to master this topic

Parent Function: Start with y = x² and identify vertical shifts
Vertex Analysis: Find y-intercept at (0, -6) to determine shift
Verification: Test point (0, -6): 0² - 6 = -6 ✓

Common Mistakes

Avoid these frequent errors

Confusing vertex position with equation form
Don't assume vertex (0, -6) means y = x² + 6 just because you see -6! This gives y-intercept of +6, not -6. Always remember: downward shift means subtracting the positive value in vertex form.

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 tell if it's +6 or -6 in the equation?

Look at where the parabola crosses the y-axis! If it crosses at y = -6, then your equation is $y = x^2 - 6$ because when x = 0, you get y = 0 - 6 = -6.

What's the difference between vertex and y-intercept?

For $y = x^2 - 6$ , the vertex is at (0, -6) and the y-intercept is also at (0, -6). This only happens when the parabola opens upward with no horizontal shift!

How do I know this isn't y = -6x²?

The graph shows a parabola opening upward, not stretched vertically. If it were $y = -6x^2$ , it would open downward and be much steeper.

Can I check my answer with another point?

Yes! Try x = 1: $y = 1^2 - 6 = -5$ . The point (1, -5) should be on your parabola. You can also try x = -1 for the same result.

Why isn't the answer y = x²/6?

That would create a horizontal stretch, making the parabola wider. The graph shows the same width as $y = x^2$ , just moved down 6 units.

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