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Put the quadratic into vertex form and state the coordinates of the vertex.

y=x^(2)+14 x+24
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+14x+24 y=x^{2}+14 x+24 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+14x+24 y=x^{2}+14 x+24 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)
  1. Identify vertex form: Identify the vertex form of a parabola.\newlineThe vertex form of a parabola is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Complete the square: Complete the square to rewrite the quadratic equation in vertex form.\newlineThe given equation is y=x2+14x+24y = x^2 + 14x + 24.\newlineTo complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial.\newlineThe coefficient of xx is 1414, so we take half of it, which is 77, and then square it to get 4949.\newlineWe add and subtract 4949 inside the equation to complete the square.
  3. Add and subtract: Add and subtract 4949 inside the equation.\newliney=x2+14x+4949+24y = x^2 + 14x + 49 - 49 + 24\newlineNow, group the perfect square trinomial and the constants.\newliney=(x2+14x+49)25y = (x^2 + 14x + 49) - 25
  4. Factor perfect square: Factor the perfect square trinomial.\newlineThe perfect square trinomial x2+14x+49x^2 + 14x + 49 can be factored into (x+7)2(x + 7)^2.\newlineSo the equation becomes:\newliney=(x+7)225y = (x + 7)^2 - 25
  5. Write in vertex form: Write the equation in vertex form and state the coordinates of the vertex.\newlineThe vertex form of the equation is y=(x+7)225y = (x + 7)^2 - 25.\newlineThe vertex (h,k)(h, k) is the point where the equation is in the form (xh)2+k(x - h)^2 + k.\newlineIn our equation, hh is 7-7 and kk is 25-25, so the vertex is (7,25)(-7, -25).

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