[tex]y=x^2 + 6x - 16[/tex]
a = 1 > 0 ⇒ Parabola opens up
[tex]y'=(x^2 + 6x - 16)'=2x+6
\\y''=(2x+6)'=2
[/tex]
y' = 2 > 0 ⇒ The vertex will be a minimum
[tex]a=1,b=6,c=-16
\\
\\T(- \frac{b}{2a} , \frac{4ac-b^2}{4a} )
\\
\\T(- \frac{6}{2\times1} , \frac{4\times1(-16)-6^2}{4\times1} )
\\
\\T(-3,-25)[/tex]
[tex]x^2 + 6x - 16=0
\\x^2+8x-2x-16=0
\\x(x+8)-2(x+8)=0
\\(x+8)(x-2)=0
\\x=-8 \text{ and } x=2
\\[/tex]
The graph intersects x-axis at -8 and 2. Its minimum is at (-3,-25), and it is opened up.
