Question

For y = x2 + 6x - 16,

Determine if the parabola opens up or down.
State if the vertex will be a maximum or minimum.
Find the vertex.
Find the x-intercepts.
Describe the graph of the equation.

Answer 1

$y=x^2 + 6x - 16$

a = 1 > 0 ⇒ Parabola opens up

$y'=(x^2 + 6x - 16)'=2x+6 \\y''=(2x+6)'=2$
y' = 2 > 0 ⇒ The vertex will be a minimum

$a=1,b=6,c=-16 \\ \\T(- \frac{b}{2a} , \frac{4ac-b^2}{4a} ) \\ \\T(- \frac{6}{2\times1} , \frac{4\times1(-16)-6^2}{4\times1} ) \\ \\T(-3,-25)$

$x^2 + 6x - 16=0 \\x^2+8x-2x-16=0 \\x(x+8)-2(x+8)=0 \\(x+8)(x-2)=0 \\x=-8 \text{ and } x=2$

The graph intersects x-axis at -8 and 2. Its minimum is at (-3,-25), and it is opened up.