Solving Quadratics
can you check if this is right if not what will it be? thanks
1) y=x^2+6x+8
y=(x+3)^2 -1
y=(x+2)(x+4)
y=0 when x=-2 and x=-6
Has a minimum y value of -4
2) y=x^2-8x+12
y=(x+4)^2 -4
y=(x-2)(x-6)
y=0 when x=2 and x=6
Has a minimum y value of -1

Question

Solving Quadratics
can you check if this is right if not what will it be? thanks
1)   y=x^2+6x+8
      y=(x+3)^2 -1
      y=(x+2)(x+4)
      y=0 when x=-2 and x=-6
Has a minimum y value of -4
2)   y=x^2-8x+12
      y=(x+4)^2 -4
      y=(x-2)(x-6)
      y=0 when x=2 and x=6
Has a minimum y value of -1

jhonyy9 · Answer

in case of first x why equal -6 ?

anonymous · Answer

positive +6

jhonyy9 · Answer

not not will be never 6

anonymous · Answer

Yeah, see basically factorize. 1. the first one can be factorized by splitting the middle term. the second one, imagine x + 3 as 1 entitiy. so you can resolve it just like x^2 - 1 is resolved. so the second eq reduces to (x+3+1)(x+3-1)

anonymous · Answer

If you can factorize, the answer comes out like a piece of cake

anonymous · Answer

Are they in correct form so far?

jhonyy9 · Answer

so than y=(x+2)(x+4)

how you solve now ?

anonymous · Answer

x^2+6x+8

jhonyy9 · Answer

how you solve after have got it factored completly like (x+2)(x+4) ?

solomonzelman · Answer

$\large\color{black}{ 1)~~~y=x^2+6x+8  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$
$\large\color{black}{ ~~~~~~~y=x^2+6x+8+1-1  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$
$\large\color{black}{ ~~~~~~~y=x^2+6x+9-1  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$
$\large\color{black}{ ~~~~~~~y=(x+3)^2-1  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$
the minimum value is at:  $\large\color{black}{ ~(-3,-1)  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$

jhonyy9 · Answer

so first x+2=0
x=-2

and x+4 =0 
so x=-4

ok ?

solomonzelman · Answer

yes, the x-intercepts are at (-2,0) and (-4,0) as it is shown above.

solomonzelman · Answer

can you find the minimum value and the x-intercepts correctly for the second problem?

solomonzelman · Answer

x-intercepts in number 2 are correct.

solomonzelman · Answer

$\large\color{black}{ 2)~~~y=x^2-8x+12  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$
$\large\color{black}{ ~~~~~~~y=x^2-8x+12+4-4  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$
$\large\color{black}{ ~~~~~~~y=x^2-8x+16-4  }$    $\LARGE\color{white}{ \rm   \left|   \right|  }$

solomonzelman · Answer

can you find the vertex?

anonymous · Answer

So are they right? i wasnt sure what the minimum of the y value was for each

anonymous · Answer

@SolomonZelman

anonymous · Answer

How do you know if it has a minimum y value of -4 or -1?