Vertex Discovery: Analyzing the Quadratic y = (x-6)² + 1

Q: Why is the vertex (6,1) and not (-6,1)?

In vertex form $ y = (x-h)^2 + k $, the vertex is (h,k). From $ (x-6)^2 + 1 $, we have h = 6 (not -6) because the expression is (x - 6).

Q: How do I remember which coordinate is which?

Think of it this way: The number inside the parentheses (after the minus sign) is your x-coordinate, and the number added or subtracted outside is your y-coordinate.

Q: What if the equation was y = (x+6)² + 1 instead?

Then you'd rewrite it as $ y = (x-(-6))^2 + 1 $, so h = -6 and the vertex would be (-6, 1). The plus sign inside means h is negative!

Q: Do I need to expand the squared term to find the vertex?

No! That would make it much harder. When the equation is already in vertex form $ y = (x-h)^2 + k $, you can read the vertex directly.

Vertex Form with Direct Identification

Find the vertex of the parabola

$y=(x-6)^2+1$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the vertex of the parabola

00:03 Use the formula to describe the parabola function

00:10 The coordinates of the vertex are (P,K)

00:14 Use this formula and find the vertex point

00:18 Substitute appropriate values according to the given data

00:21 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the vertex of the parabola

$y=(x-6)^2+1$

Step-by-step solution

To find the vertex of the parabola given by the equation $y = (x-6)^2 + 1$ , we observe that this equation is already in the vertex form:

$y = (x-h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

From the equation $y = (x-6)^2 + 1$ , we can clearly identify:

$h = 6$ , because the expression inside the square is $(x-6)$ .
$k = 1$ , the constant added to the squared term.

Therefore, the vertex of the parabola is $(6, 1)$ .

Comparing this with the available choices, we see that choice 4, $(6, 1)$ , is the correct answer.

In conclusion, the vertex of the parabola is $\boxed{(6, 1)}$ .

Final Answer

$(6,1)$

Key Points to Remember

Essential concepts to master this topic

Rule: Vertex form y = (x-h)² + k gives vertex (h,k)
Technique: From (x-6)² + 1, identify h = 6 and k = 1
Check: Substitute x = 6: y = (6-6)² + 1 = 0 + 1 = 1 ✓

Common Mistakes

Avoid these frequent errors

Confusing the sign of h in vertex form
Don't think (x-6)² gives h = -6! The negative inside the parentheses means the opposite: h = 6. Always remember that y = (x-h)² + k has vertex at (h,k), so (x-6) means h = +6.

Practice Quiz

Test your knowledge with interactive questions

The following function has been plotted on the graph below:

$$ f(x)=x^2-8x+16 $$

Calculate point C.

FAQ

Everything you need to know about this question

Why is the vertex (6,1) and not (-6,1)?

In vertex form $y = (x-h)^2 + k$ , the vertex is (h,k). From $(x-6)^2 + 1$ , we have h = 6 (not -6) because the expression is (x - 6).

How do I remember which coordinate is which?

Think of it this way: The number inside the parentheses (after the minus sign) is your x-coordinate, and the number added or subtracted outside is your y-coordinate.

What if the equation was y = (x+6)² + 1 instead?

Then you'd rewrite it as $y = (x-(-6))^2 + 1$ , so h = -6 and the vertex would be (-6, 1). The plus sign inside means h is negative!

Do I need to expand the squared term to find the vertex?

No! That would make it much harder. When the equation is already in vertex form $y = (x-h)^2 + k$ , you can read the vertex directly.

How can I check if my vertex is correct?

Substitute the x-coordinate into the equation. At the vertex, the squared term $(x-h)^2$ equals zero, so y should equal k. For (6,1): $(6-6)^2 + 1 = 0 + 1 = 1$ ✓

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Vertex Discovery: Analyzing the Quadratic y = (x-6)² + 1

Vertex Form with Direct Identification

❤️ 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

Why is the vertex (6,1) and not (-6,1)?

How do I remember which coordinate is which?

What if the equation was y = (x+6)² + 1 instead?

Do I need to expand the squared term to find the vertex?

How can I check if my vertex is correct?

🌟 Unlock Your Math Potential

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