Question

X^2 + 20x + 28 = 9 find the graphing/vertex form

Answer 1

Answer:
`(x+10)^2+119=0`

Explanation:

For a quadratic equation, the vertex form is given by :
`y=a(x-h)^2+k`, where (h, k) is the vertex.

The given quadratic equation:
`x^2 + 20x + 28 = 9`

Subtract 9 from both sides

`=x^2+20x+19=0`

compare this to
`x^2+bx=c`, and add
`((b)/(2))^2` both sides

b= 20

`x^2+20+100+19=-100` [(b/2)²=20/2=10]

`\Rightarrow\ x^2+2(x)(10)+10^2+119=0`

`\Rightarrow\ (x+10)^2+119=0\ \ \ [\because\ (a+b)^2=a^2+b^2+2ab]`

So, the vertex form :
`(x+10)^2+119=0`