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.
Show all work and use complete sentences

Answer 1

We can find the vertex first
\[y=x^2+6x-16\]\[y+16=x^2+6x\] now we'll be completing the square . \[y+16=x^2+\color{red}6x\]\[y+16+\color{red}9=x^2+6x+\color{red}{9}\]\[y+25=(x+3)^2\]\[y=(x+3)^2-25\]\[y=(x-(\color{blue}{-3}))+(\color{blue}{-25})\]
Your vertex.

Answer 2

we know from the previous it ups upward

Answer 3

*opens

Answer 4

x intercept when y equals 0

Answer 5

it's simple. If the co-eff of x^2 is positive, it opens up, if negative, opens down.

Answer 6

we could use the first derivative test to determine if it's a max or min.. since it opens up it's a min..

Answer 7

The formula \[\large y=\color{purple}a(x-h)^2+k\] is the formula for a quadratic, and your "a" value tells if it opens up or down,
your formula \[\large y=\color{purple}{(1)}(x-(-3))^2-25\] the "a" is positive, so your parabola opens upward.

Answer 8

@Jhannybean good job.

Answer 9

It's her thing :P

Answer 10

you can find your x-intercept by setting y=0. \[\large (x+3)^2-25=0\]\[\large (x+3)^2=25\]\[\large \sqrt{(x+3)^2}= \pm \sqrt{25}\]\[\large |x+3|= \pm \sqrt{25}\] solving for x would give you your x-intercepts. \[\large |x+3| = \pm 5\]

Answer 11

0=(x+8)(x-2)

Answer 12

When describing the graph of the equation, you could use your min to state where the slope is inc/dec, and the domain/range of the function

Answer 13

@sumsumjoe we're not doing all this work for you if you're not gonna try and understand..

Answer 14

Well Jhan won't :P

Answer 15

^_^

Answer 16

and as always.. great job :D

Answer 17

Lol, thanks

Answer 18

I think he's gone :l