How do i put y=x^2+8x+14 into vertex form

Question

That’s it

Answer 1

y = x²  +8x + 14

 

To put the equation into vertex form, we are going to "complete the square" on the right hand side of the equation:

 

y = (x² + 8x + C) + 14 - C

 

We are going to add a number C to the first two terms that complete the square.  Since we added C to the RHS, we need to subtract it again so we don't change anything (C - C = 0).  The value of C is half of the coefficient of the x term (8) squared, that is C = (8/2)² = 4² = 16.  So our equation is:

 

y = (x² + 8x + 16) + 14 - 16 =  (x² + 8x + 16) - 2

 

Now x² + 8x + 16 is a perfect square equal to x + half of the coefficient of the x term (8), all squared:

 

x² + 8x + 16 = (x + 8/2)² = (x + 4)²

 

Our final equation, in vertex form, is:

 

y = (x+4)² - 2

 

Answer 2

y=x^2+8x+14

y-14=x^2+8x

complete the square

[(1/2)*(8)]^2=16

y-14+16=x^2+8x+16

y+2=(x+4)^2

y=(x+4)^2-2