Question

For y = x^2 + 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.

Show all work and use complete sentences to receive full credit.

Answer 1

i think its an up and minimum but that's all i know

Answer 2

http://openstudy.com/study#/updates/4f871838e4b0505bf086b15d

Answer 3

anytime a is greater than 0 in a quadratic the parabola is going to open 'up'. If a is negative it will be reflected across the x axis and open 'down'

Answer 4

tyvm

Answer 5

and yes it is a minimum

Answer 6

the rest of it should be on the link nurali provided

Answer 7

but just incase y = x^2 + 6x − 16
y + 9 = (x^2 + 6x + 9) -16
y+9=(x+3)^2-16
y=(x+3)^2- 25
so the vertex is going to be -3,-25

Answer 8

and what i just did is called completing the square. you take the b value(6) divide it by 2, then square it and add it to both sides.

Answer 9

then group and factor the new quadratic like normal, and subtract the new number from both sides to isolate y

Answer 10

the x intercepts will 8 and -2 since y = x^2 + 6x − 16 = y = (x+8)(x-2)

Answer 11

to find those you just factor the original quadratic like usual. i don't really understand what is meant by describe the graph but i think they mean that the graph is translated 3 units to the left and 25 units down