Question

The vertex of the parabola y = x2 + 8x + 10 lies in Quadrant

Answer 1

`y = x^2 + 8x + 10 \\ \\the \ standard \ form \ y = ax^2 + bx + c \\\\of \ a \ function \ into \ vertex \ form \ y = a(x - h)^2 + k ,\\ \\ we \ have \ to \ write \ the \ equation \ in \ the \ complete \ square \ form \\\\\ and \ vertex(h, k) \ is \ given \ by:`


`h = (-b)/(2a) , \ \ k = c -(b^2)/(4a ) \\ \\y = a(x - h)^2+k`

opens up for a > 0


`a=1 , \ \ b=8, \ \ c=10 \\ \\h= (-8)/(2)=-4`


`k= 10-(8^2)/(4)=10-(64)/(4)=10-16=-6 \\ \\y=(x-(-4))^2+(-6)\\ \\y=(x+4)^2-6`

Answer 2

`y = x^2 + 8x + 10\\\\The\ vertex=(p;\ q)\ \ \ and\ \ \ p=- (b)/(2a) ;\ \ \ q=- (\Delta)/(4a) ;\ \ \ \Delta=b^2-4ac\\\\\Delta=8^2-4\cdot1\cdot10=64-40=24\\\\p=- (8)/(2\cdot1) =-4\\\\q=- (24)/(4\cdot1) =-6\\\\the\ vertex=(-4;-6)`