For y = x2 + 4x − 12,

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.

Question

For y = x2 + 4x − 12,

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.

wolf1728 · Answer

The coefficient of x² is positive so it is 'U' shaped and opens up.

anonymous · Answer

$$y=x ^{2}+4x+4-4-12$$
$$y+16=\left( x+2 \right)^{2},\it is an upward parabola with vertex (-2,-16)$$
It is minimum at the vertex.
for x-intercepts put y=0
$$x+2=\pm 4,x=4-2=2 and x=-4-2=-6$$