Find the Vertex of the Parabola: (x-1)²-1

Vertex Form with Direct Identification

Find the vertex of the parabola

y=(x−1)2−1 y=(x-1)^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 We'll use the formula to describe a parabolic function
00:07 The coordinates of the vertex are (P,K)
00:11 We'll use this formula to find the vertex point
00:18 We'll substitute appropriate values according to the given data
00:21 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the vertex of the parabola

y=(x−1)2−1 y=(x-1)^2-1

2

Step-by-step solution

The given equation is y=(x−1)2−1 y = (x-1)^2 - 1 . This equation is in the vertex form, y=a(x−h)2+k y = a(x-h)^2 + k , where a a , h h , and k k are constants.

In this case, the given equation can be written as y=1⋅(x−1)2+(−1) y = 1 \cdot (x-1)^2 + (-1) , indicating that a=1 a = 1 , h=1 h = 1 , and k=−1 k = -1 .

The vertex form of the quadratic equation allows us to directly identify the vertex of the parabola as (h,k) (h, k) .

From our identification, it is clear that the vertex (h,k) (h, k) of the parabola is (1,−1) (1, -1) .

Therefore, the vertex of the given parabola is (1,−1) (1, -1) .

3

Final Answer

(1,−1) (1,-1)

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: y=a(x−h)2+k y = a(x-h)^2 + k gives vertex at (h,k) (h,k)
  • Technique: From y=(x−1)2−1 y = (x-1)^2 - 1 , identify h=1 h = 1 and k=−1 k = -1
  • Check: Substitute x=1 x = 1 : y=(1−1)2−1=0−1=−1 y = (1-1)^2 - 1 = 0 - 1 = -1 ✓

Common Mistakes

Avoid these frequent errors
  • Confusing the signs in vertex form
    Don't read (x−1)2 (x-1)^2 as h = -1! The vertex form is (x−h)2 (x-h)^2 , so (x−1)2 (x-1)^2 means h = 1. Always remember: if it's (x−1) (x-1) , then h = 1; if it's (x+1) (x+1) , then h = -1.

Practice Quiz

Test your knowledge with interactive questions

The following function has been graphed below:

\( f(x)=x^2-6x \)

Calculate point C.

CCCAAABBB

FAQ

Everything you need to know about this question

How do I know which number is h and which is k?

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In vertex form y=a(x−h)2+k y = a(x-h)^2 + k , h is the number inside the parentheses (with opposite sign) and k is the number added or subtracted at the end.

What if the equation doesn't look like vertex form?

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If you have y=ax2+bx+c y = ax^2 + bx + c , you need to complete the square first to convert it to vertex form. But this problem is already in vertex form!

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

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From y=(x−1)2−1 y = (x-1)^2 - 1 : the 1 inside the parentheses gives h = 1, and the -1 outside gives k = -1. So vertex is (1,-1).

How can I double-check my vertex is correct?

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Substitute the x-coordinate into the equation! At x = 1: y=(1−1)2−1=0−1=−1 y = (1-1)^2 - 1 = 0 - 1 = -1 . This confirms the vertex is (1,-1).

Does the 'a' value affect the vertex location?

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No! The value of 'a' only affects how wide or narrow the parabola is and whether it opens up or down. The vertex position depends only on h and k values.

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