Find the Vertex of y=(x+2)²-2: Quadratic Function Analysis

Vertex Form with Sign Identification

Find the vertex of the parabola

y=(x+2)22 y=(x+2)^2-2

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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 the parabola function
00:08 The coordinates of the vertex are (P,K)
00:12 We'll use this formula and find the vertex point
00:16 We notice that according to the formula, the term P is negative
00:27 We'll substitute appropriate values according to the given data
00:30 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+2)22 y=(x+2)^2-2

2

Step-by-step solution

To solve the problem of finding the vertex of the parabola represented by y=(x+2)22 y = (x+2)^2 - 2 , consider the form of the equation.

This equation is written in the vertex form of a quadratic function, which is given by:

y=a(xh)2+k y = a(x-h)^2 + k

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

Comparing y=(x+2)22 y = (x+2)^2 - 2 with the standard form y=a(xh)2+k y = a(x-h)^2 + k , we identify:

  • h=2 h = -2 because the expression is (x(2))2 (x - (-2))^2
  • k=2 k = -2

Therefore, the vertex of the parabola is (2,2)(-2, -2).

The correct answer is (2,2)(-2, -2).

3

Final Answer

(2,2) (-2,-2)

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: In y=a(xh)2+k y = a(x-h)^2 + k , vertex is always (h,k) (h, k)
  • Sign Rule: For (x+2)2 (x+2)^2 , rewrite as (x(2))2 (x-(-2))^2 so h = -2
  • Verify: Substitute x = -2: y=(2+2)22=02=2 y = (-2+2)^2 - 2 = 0 - 2 = -2

Common Mistakes

Avoid these frequent errors
  • Misidentifying the h-value from (x+2)²
    Don't think h = +2 just because you see +2 in the parentheses = wrong vertex (2,-2)! The vertex form is (x-h)², so (x+2)² means (x-(-2))², making h = -2. Always rewrite (x+number)² as (x-(-number))² to find h correctly.

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

Why is the x-coordinate negative when I see +2 in the equation?

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The vertex form is y=a(xh)2+k y = a(x-h)^2 + k with a minus sign before h. When you see (x+2)2 (x+2)^2 , think of it as (x(2))2 (x-(-2))^2 , so h = -2.

How do I remember which coordinate is which?

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Use the memory trick: h comes before k in the alphabet, and x comes before y in coordinates. So vertex = (h, k) = (x-coordinate, y-coordinate).

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

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

Can I check my vertex answer somehow?

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Yes! Substitute the x-coordinate back into the equation. You should get the y-coordinate. For (-2, -2): y=(2+2)22=02=2 y = (-2+2)^2 - 2 = 0 - 2 = -2

Does the 'a' value affect the vertex location?

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No! The coefficient 'a' only affects how wide or narrow the parabola is and whether it opens up or down. The vertex (h, k) stays the same regardless of 'a'.

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