Question

How do i graph this function?
y = -(x+4) - 1

also show how to do it

Answer 1

First put the equation in slope-intercept form.
y = -x -4 -1
y =-x -5
Slope is -1/1 and the y intercept is -5.

Answer 2

lol too much work :P

Answer 3

Surely, if she is asking how to do it here, she is expected to be able to do it without technology, don't you think?

Answer 4

Plot the y intercept on the graph and make a line using the given slope, (-1/1). down 1 right 1 or up 1 left 1.

Answer 5

If you are doing transformations of functions, then it might be expected that you would do it with transformations.

This is a transformation of the equation y=x.

But if not, then @shamil98 's method is perfect.

Answer 6

were supposed to make a parabola lol im so confused but would i plot -1 and then do -1/1

Answer 7

There are no mentions of trigonometric functions (e.g. sin , cos, tan, cotan, etc.) so I assumed it would be this method?.. but if it was it would me different with phase shifts and vertical shifts.

Answer 8

Then you gave us the wrong equation, lol.

Is the stuff in the ( )'s squared??

Answer 9

yeah its squared thats exactly how the problem is

Answer 10

Oh, a parabola. My bad.

Answer 11

@shamil98 I teach transformations in algebra 2, no trig required.

Answer 12

If that's the equation you are given, that is NOT a parabola.

If you are given \(y = -(x+4)^2 - 1\) then THAT is a parabola.

Answer 13

The squared part should be shown as ^2 , for future reference , helps us helpers know what to do. lol.

Answer 14

yeah sorry its the one you just said @DebbieG

Answer 15

ok, if it's \(y = -(x+4)^2 - 1\), then again, it kind of depends on what you've done in your class.

If you have done transformations of functions, I would explain it one way.

If you haven't but are just being introduced to graphing quadratic equations, then I would explain it a bit differently... so I'll assume the latter.

Answer 16

\(y = -(x+4)^2 - 1\) is called "vertex form", because you can read the vertex right off the equation:

\(y = a(x-h)^2 +k\) has vertex (h,k) Does that look familiar?

So what is the vertex of YOUR equation, \(y = -(x+4)^2 - 1\) ?

Answer 17

xy i think ??

Answer 18

Once you have the vertex, you can look at the sign of a to decide it it opens up or opens down.

If a>0, opens up
if a<0, opens down.

Answer 19

The vertex is a POINT, so you should have a coordinate pair, (x, y)

The question is, what is x, and what is y?

Answer 20

Look at \(y = a(x-h)^2 +k\)

and compare it to your equation,

\(y = -(x+4)^2 - 1\)

Find h and k. this might help:

\(y = -(x-\color{red}{(-4)})^2 +\color{red}{(- 1)}\)
\(y = a(x-\color{red}h)^2 +\color{red}k\)

Answer 21

so would you fill in the x and y slots with the -4 and -1

Answer 22

Yes, that gives you the point that is the vertex. (-4, -1) is the vertex.

Does this sound familiar? This should be what you are covering in this class, given that you are asked this problem

Answer 23

You know what the "vertex" of a parabola is, right?

Answer 24

well i missed the last 3 days so im behind and the teacher didnt really help me catch up so i dont know most of this lol

Answer 25

OK, well, at a minimum you should sit down and read your textbook before trying to jump into the problems; and get class notes from someone. But we can move forward, as long as you are understanding, so far.

Answer 26

OK, so we have the vertex. Now, which direction does this parabola open - up or down? (I talked about that above, scroll up if you need to read it again)

Answer 27

down

Answer 28

Very good. so we have:

\(y = -(x+4)^2 - 1\) has vertex at (-4,-1) and opens downward.

Now we want to see if we can figure out a couple other things to help us with the graph.

How about the y-intercept? How do we find that?

Answer 29

Just like for a line or another graph, the y-intercept happens when ___ =0?

What goes in the ___ ? :)

Answer 30

when x = 0 i think right ? im sorry im terrible at math

Answer 31

Yes, when x=0.

Don't ever say that you're terrible at math... you've said that your whole life and now you're convinced of it, aren't you? :)

Now we want to find the value of that y-intercept, and we know it happens when x=0.

So we just plug in x=0 into our equation:

\(y = -(x+4)^2 - 1\)

so we have

\(y = -(0+4)^2 - 1=???\)

Answer 32

can you evaluate that and tell me what that y-value will be?

Answer 33

15 ?

Answer 34

oops,close, but the - sign in front of the ( )'s happens AFTER you square the 4, right?

Answer 35

\(y = -(4)^2 - 1= -16 - 1=??\)

Answer 36

oh so -17

Answer 37

See, it ISN'T \((-4)^2\), it's \(-(4)^2\), that's the difference.

RIGHT, -17.

Answer 38

OK, now we have the vertex, we know it opens downward and we know that the y-intercept is at -17. That's quite a bit of info so far:

Answer 39

Now, the next thing that we would usually do is to find the x-intercepts... which is where y=0 (where the graph crosses the x-axis.

Only, with this function, there is one problem with that. Do you see what it is?

Why am I not going to look for x-intercepts?

Answer 40

i have no clue lol

Answer 41

lol.. ok, look at some different parabolas:


Notice that one of the downward opening ones has x-intercepts, the other doesn't.

Notice that one of the upward opening ones has x-intercepts, the other doesn't.

Do you see what makes the difference?