Find the Vertex of y=(x-5)²+1: Parabola Analysis

Q: Why is the vertex (5,1) and not (-5,1)?

In vertex form $ y = a(x-h)^2 + k $, the vertex is at (h,k). When you see $ (x-5)^2 $, this means h = 5 because we're subtracting 5 from x. The negative sign is already built into the form!

Q: What if the equation was y = (x+5)² + 1 instead?

Then the vertex would be (-5, 1)! The expression $ (x+5)^2 $ can be written as $ (x-(-5))^2 $, so h = -5. Remember: plus in the parentheses means negative h-value.

Q: How do I remember which coordinate is which?

Think of it as (h,k) = (horizontal shift, vertical shift). The h-value moves the parabola left or right, and the k-value moves it up or down from the origin.

Q: Do I need to expand the equation to find the vertex?

No! That's the beauty of vertex form - the vertex is immediately visible. Expanding would just make your work harder and more prone to errors.

Vertex Form Analysis with Standard Parameters

Find the vertex of the parabola

$y=(x-5)^2+1$

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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

00:20 Substitute appropriate values according to the given data

00:23 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 vertex of the parabola

$y=(x-5)^2+1$

Step-by-step solution

To solve this problem, we will determine the vertex of the parabola from the equation $y = (x-5)^2 + 1$ , which is in vertex form.

Step 1: Recognize the form of the quadratic.
The given equation is $y = (x-5)^2 + 1$ , resembling the vertex form $y = a(x-h)^2 + k$ .
Step 2: Identify the vertex parameters $(h, k)$ .
In the equation $y = (x-5)^2 + 1$ :
$h = 5$ and $k = 1$ .
Step 3: Write down the coordinates of the vertex based on the identified values.
Therefore, the vertex of the parabola is at the point $(5, 1)$ .

Thus, the vertex of the parabola is $(5, 1)$ .

Final Answer

$(5,1)$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = a(x-h)^2 + k$ directly shows vertex coordinates
Parameter Identification: From $y = (x-5)^2 + 1$ , h = 5 and k = 1
Verification: Substitute x = 5: $y = (5-5)^2 + 1 = 1$ gives point (5,1) ✓

Common Mistakes

Avoid these frequent errors

Confusing the sign of h in vertex form
Don't write the vertex as (-5,1) when you see (x-5)! The expression (x-5) means h = +5, not -5. The vertex form y = a(x-h)² + k uses subtraction, so (x-5) gives h = 5. Always remember: the vertex is at (h,k) where h has the opposite sign of what appears in the parentheses.

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 (5,1) and not (-5,1)?

In vertex form $y = a(x-h)^2 + k$ , the vertex is at (h,k). When you see $(x-5)^2$ , this means h = 5 because we're subtracting 5 from x. The negative sign is already built into the form!

What if the equation was y = (x+5)² + 1 instead?

Then the vertex would be (-5, 1)! The expression $(x+5)^2$ can be written as $(x-(-5))^2$ , so h = -5. Remember: plus in the parentheses means negative h-value.

How do I remember which coordinate is which?

Think of it as (h,k) = (horizontal shift, vertical shift). The h-value moves the parabola left or right, and the k-value moves it up or down from the origin.

Do I need to expand the equation to find the vertex?

No! That's the beauty of vertex form - the vertex is immediately visible. Expanding would just make your work harder and more prone to errors.

What does the coefficient 'a' tell me?

The coefficient a determines the parabola's direction and width. Since this equation has no visible coefficient, a = 1, meaning the parabola opens upward with standard width.

How can I check my vertex is correct?

Substitute the x-coordinate back into the equation. For vertex (5,1): $y = (5-5)^2 + 1 = 0 + 1 = 1$ . The y-value matches, confirming our vertex is correct!

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Find the Vertex of y=(x-5)²+1: Parabola Analysis

Vertex Form Analysis with Standard Parameters

❤️ 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 (5,1) and not (-5,1)?

What if the equation was y = (x+5)² + 1 instead?

How do I remember which coordinate is which?

Do I need to expand the equation to find the vertex?

What does the coefficient 'a' tell me?

How can I check my vertex is correct?

🌟 Unlock Your Math Potential

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