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

y=x^(2)+10 x+24
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+10x+24 y=x^{2}+10 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+10x+24 y=x^{2}+10 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.\newlineGiven equation: y=x2+10x+24y = x^2 + 10x + 24\newlineTo complete the square, we need to find a number that, when added and subtracted to the equation, forms a perfect square trinomial with x2+10xx^2 + 10x.\newlineHalf of the linear coefficient (1010) is 55, and squaring it gives us 2525.\newlineSo, we add and subtract 2525 inside the equation to complete the square.\newliney=x2+10x+2525+24y = x^2 + 10x + 25 - 25 + 24
  3. Rewrite equation and combine: Rewrite the equation with the perfect square trinomial and combine the constants.\newliney=(x2+10x+25)1y = (x^2 + 10x + 25) - 1\newlineNow, factor the perfect square trinomial.\newliney=(x+5)21y = (x + 5)^2 - 1\newlineThis is the vertex form of the equation.
  4. Identify vertex: Identify the vertex of the parabola.\newlineFrom the vertex form y=(x+5)21y = (x + 5)^2 - 1, we can see that the vertex (h,k)(h, k) is (5,1)(-5, -1).

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