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y=(x-2)(x+4)
The given equation represents a parabola in the 
xy-plane. Which of the following equivalent forms of the equation displays the coordinates of the vertex of the parabola as constants or coefficients?
Choose 1 answer:
(A) 
y=x^(2)+2x-8
(B) 
y=(x+1)^(2)-9
(c) 
y-40=(x-6)(x+8)^(')
(D) 
y+5=(x-1)(x+3)

y=(x2)(x+4) y=(x-2)(x+4) \newlineThe given equation represents a parabola in the xy x y -plane. Which of the following equivalent forms of the equation displays the coordinates of the vertex of the parabola as constants or coefficients?\newlineChoose 11 answer:\newline(A) y=x2+2x8 y=x^{2}+2 x-8 \newline(B) y=(x+1)29 y=(x+1)^{2}-9 \newline(C) y40=(x6)(x+8) y-40=(x-6)(x+8) \newline(D) y+5=(x1)(x+3) y+5=(x-1)(x+3)

Full solution

Q. y=(x2)(x+4) y=(x-2)(x+4) \newlineThe given equation represents a parabola in the xy x y -plane. Which of the following equivalent forms of the equation displays the coordinates of the vertex of the parabola as constants or coefficients?\newlineChoose 11 answer:\newline(A) y=x2+2x8 y=x^{2}+2 x-8 \newline(B) y=(x+1)29 y=(x+1)^{2}-9 \newline(C) y40=(x6)(x+8) y-40=(x-6)(x+8) \newline(D) y+5=(x1)(x+3) y+5=(x-1)(x+3)
  1. Expand and Match: To find the vertex form of the equation, we need to complete the square. The vertex form of a parabola's equation is y=a(xh)2+ky=a(x-h)^2+k, where (h,k)(h,k) is the vertex of the parabola.
  2. Convert to Vertex Form: First, let's expand the given equation y=(x2)(x+4)y=(x-2)(x+4) to see if it matches any of the answer choices.\newliney=x2+4x2x8y = x^2 + 4x - 2x - 8\newliney=x2+2x8y = x^2 + 2x - 8\newlineThis matches answer choice (A), but this form does not show the vertex.
  3. Check Answer Choices: Now, let's try to convert the expanded form into the vertex form by completing the square.\newlineWe have y=x2+2x8y = x^2 + 2x - 8.\newlineTo complete the square, we take the coefficient of xx, which is 22, divide it by 22, and square it. (2/2)2=12=1(2/2)^2 = 1^2 = 1.\newlineWe add and subtract this number inside the equation to complete the square.\newliney=(x2+2x+1)18y = (x^2 + 2x + 1) - 1 - 8\newliney=(x+1)29y = (x + 1)^2 - 9\newlineThis matches answer choice (B) and shows the vertex as (1,9)(-1, -9).
  4. Final Answer: None of the other answer choices are in vertex form, so we do not need to check them further. Answer choice (B) is the correct answer because it is in vertex form and displays the coordinates of the vertex of the parabola as constants or coefficients.

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