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5 votes
Y=x2^+9x+7 in vertex form

User Birish
by
4.9k points

2 Answers

2 votes

Answer: (-9/2,-53/4)

Explanation:

User Mockinterface
by
5.3k points
4 votes

Answer:


y=\left(x+(9)/(2)\right)^2-(53)/(4)

Explanation:

Vertex form of a quadratic equation:


\boxed{y=a(x-h)^2+k}

where:

  • (h, k) is the vertex.
  • a is the leading coefficient.

To find the vertex form of a quadratic equation, complete the square.

Given quadratic equation:


y=x^2+9x+7

Add and subtract the square of half the coefficient of the term in x to the right side of the equation:


\implies y=x^2+9x+\left((9)/(2)\right)^2+7-\left((9)/(2)\right)^2


\implies y=x^2+9x+(81)/(4)+7-(81)/(4)


\implies y=x^2+9x+(81)/(4)-(53)/(4)

Factor the perfect square trinomial formed by the first 3 terms:


\implies y=\left(x+(9)/(2)\right)^2-(53)/(4)

User HcgRandon
by
5.8k points