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

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

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Lilith Lilith · Answer 2 · 2014-06-19T15:57:44+02:00

[tex]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:[/tex]

[tex]h = \frac{-b}{2a} , \ \ k = c -\frac{b^2}{4a } \\ \\y = a(x - h)^2+k[/tex]

opens up for a > 0

[tex] a=1 , \ \ b=8, \ \ c=10 \\ \\h= \frac{-8}{2}=-4[/tex]

[tex]k= 10-\frac{8^2}{4}=10-\frac{64}{4}=10-16=-6 \\ \\y=(x-(-4))^2+(-6)\\ \\y=(x+4)^2-6[/tex]