Identify the Vertex in the Quadratic Equation: y = (x - 5) - 5

Vertex Form with Missing Squared Terms

Find the vertex of the parabola

y=(x5)5 y=(x-5)-5

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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
00:07 The coordinates of the vertex are (P,K)
00:12 Use this formula and find the vertex point
00:16 Substitute appropriate values according to the given data
00:20 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=(x5)5 y=(x-5)-5

2

Step-by-step solution

To solve the problem of finding the vertex of the parabola given by y=(x5)5 y = (x - 5) - 5 , we start by recognizing that the equation is not yet in the vertex form y=a(xh)2+k y = a(x-h)^2 + k . This format directly shows the vertex (h,k)(h, k).

The given equation is y=(x5)5 y = (x - 5) - 5 . First, we should reinterpret the equation in a way that reflects the standard parabolic structure. This is rewritten as:

y=x55 y = x - 5 - 5 .

This simplifies to:

y=x10 y = x - 10 .

It's important to realize here that the equation appears linear due to simplification; hence it does not perfectly fit conventional expectations for a parabola in vertex form y=a(xh)2+k y = a(x-h)^2 + k . However, for quadratic functions, the standard formulation involves a squared term, which is missing here, indicating a potential typographical or conceptual error in format.

Given the choices, align (xh)2 (x - h)^2 assuming its effects shown as horizontal and vertical shifts around an understood correction if it should indeed equate to linear features. Often a restatement to fully align might be required where h=5 h = 5 and k=5 k = -5 relating to placeholders assumed within broader expectations for concise interpretation as a quadratic-continuous or misaddressed.

Thus, the vertex consistent with expectations would be:

(5,5) (5, -5) , matching choice 1.

Therefore, the vertex of the parabola is (5,5) (5, -5) .

3

Final Answer

(5,5) (5,-5)

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: Standard form is y=a(xh)2+k y = a(x-h)^2 + k with vertex (h,k) (h,k)
  • Pattern Recognition: From y=(x5)5 y = (x-5) - 5 , identify h = 5 and k = -5
  • Verification: Check that vertex (5,5) (5,-5) matches the horizontal and vertical shifts ✓

Common Mistakes

Avoid these frequent errors
  • Treating incomplete equations as standard vertex form
    Don't assume y=(x5)5 y = (x-5) - 5 is complete vertex form = wrong interpretation! The missing squared term suggests either a typo or misunderstanding. Always recognize that true vertex form requires (xh)2 (x-h)^2 and identify the intended pattern from context.

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 doesn't this equation look like a normal parabola?

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You're right to notice something odd! The equation y=(x5)5 y = (x-5) - 5 is missing the squared term that makes it truly quadratic. This appears to be either a typo or an incomplete form.

How can I find the vertex if the equation isn't complete?

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Look for the pattern in vertex form. Even with missing parts, (x5) (x-5) suggests h = 5 (horizontal shift), and the -5 suggests k = -5 (vertical shift), giving vertex (5,5) (5,-5) .

What if this was supposed to be a squared term?

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If it should be y=(x5)25 y = (x-5)^2 - 5 , then the vertex is still (5,5) (5,-5) ! The vertex coordinates come from the values inside and outside the parentheses.

Is this actually a linear equation instead?

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When simplified, y=(x5)5=x10 y = (x-5) - 5 = x - 10 is indeed linear. However, given the context and answer choices, this problem expects you to identify the intended vertex pattern.

How do I remember which number goes with which coordinate?

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In y=a(xh)2+k y = a(x-h)^2 + k , the vertex is (h,k) (h,k) . Remember: h comes from inside the parentheses (opposite sign), and k is the constant added outside.

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