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Rewrite the function by completing the square. {:[f(x)=x^(2)+2x-39],[f(x)=(x+◻)^(2)+◻]:}

Rewrite the function by completing the square.\newlinef(x)=x2+2x−39 f(x) = x^2 + 2x - 39 \newlinef(x)=(x+□)2+□ f(x) = (x + \square)^2 + \square

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Q. Rewrite the function by completing the square.\newlinef(x)=x2+2x−39 f(x) = x^2 + 2x - 39 \newlinef(x)=(x+□)2+□ f(x) = (x + \square)^2 + \square
  1. Given quadratic function: We start with the given quadratic function:\newlinef(x) = x2+2x−39x^2 + 2x - 39\newlineOur goal is to rewrite this function in the form of (x+p)2+q(x + p)^2 + q, where pp and qq are constants.
  2. Completing the square: To complete the square, we need to find a value that, when added and subtracted to the x2+2xx^2 + 2x part, forms a perfect square trinomial.\newlineThe coefficient of xx is 22, so we take half of it, which is 11, and then square it to get 12=11^2 = 1.
  3. Adding and subtracting the value: We add and subtract this value inside the function to create a perfect square trinomial:\newlinef(x) = (x2+2x+1)−1−39(x^2 + 2x + 1) - 1 - 39
  4. Simplifying the constant terms: Now we simplify the constant terms: f(x)=(x2+2x+1)−40f(x) = (x^2 + 2x + 1) - 40
  5. Factoring the perfect square trinomial: The expression x2+2x+1x^2 + 2x + 1 is a perfect square trinomial and can be factored as (x+1)2(x + 1)^2:\newlinef(x)=(x+1)2−40f(x) = (x + 1)^2 - 40
  6. Final answer: We have now rewritten the function in the completed square form:\newlinef(x) = (x + 11)^22 - 4040\newlineThis is the final answer.

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