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

Answer 1

The first two should be simple, which is to determine if the parabola opens up or down and if it will be a maximum or minimum. What do you think? Does it open up or down and whats the maximum or minimum? Here is a hint. If a is negative it opens one way and if a is positive it opens another way. Once you know how the parabola opens up, you will be able to state if the vertex will be a max or mini. If the vertex opens up. The vertex will be a minimum. If the vertex opens down. The vertex will be a maximum.

Answer 2

the coefficient of x^2 is 1 so the parabola is concave up

the vertex will be a minimum since the parabola is concave up.

Vertex... there are several methods..

use the line of symmetry \[x = \frac{-b}{2a}\]

in your question a = 1 and b = 4

substitute this value onto the original equation to find y... this will be the ordered pair for the vertex.

x intercepts occur when y = 0

so you need to solve

\[x^2 + 4x - 12 = 0\]

this can be factorised....

hope this helps

Answer 3

@Migitmack can you follow @campbell_st's instruction?

Answer 4

a little

Answer 5

it kinda confusing

Answer 6

i just don't understande last part

Answer 7

Describe the graph of the equation.

Answer 8

what does that mean

Answer 9

you need to sketch the graph using the information above

that would be my best guess