Question

Describe the graph of the function f(x) = x3 − 15x2 + 68x − 96. Include the y-intercept, x-intercepts, and the shape of the graph.

Answer 1

Someone please help

Answer 2

If you replace every x with 0, what is the result?

In other words, what is the value of f(0)?

Answer 3

0.. im not really sure how to do any of it.

Answer 4

\[\Large f(x) = x^3 - 15x^2 + 68x - 96\]
\[\Large f({\color{red}{x}}) = ({\color{red}{x}})^3 - 15({\color{red}{x}})^2 + 68({\color{red}{x}}) - 96\]
\[\Large f({\color{red}{0}}) = ({\color{red}{0}})^3 - 15({\color{red}{0}})^2 + 68({\color{red}{0}}) - 96\]
\[\Large f(0) = ???\]
Use PEMDAS to simplify. Tell me what you get.

Answer 5

-96?

Answer 6

correct

Answer 7

when x = 0, it leads to y = -96

Answer 8

so (x,y) = (0,-96) is the y intercept as an ordered pair. This is where it crosses the y axis.

Answer 9

make sense?

Answer 10

Yes a little bit. So is that how i would fully graph it? like what do i do for that?

Answer 11

Is your teacher letting you use a graphing calculator or technology to graph this? or does your teacher want you to do it by hand?

Answer 12

by hand.

Answer 13

ok so what you'll need to do is generate a table of points

Answer 14

(0,-96) is one point. Let's find another

when x = 1, then

\[\Large f(x) = x^3 - 15x^2 + 68x - 96\]
\[\Large f({\color{red}{x}}) = ({\color{red}{x}})^3 - 15({\color{red}{x}})^2 + 68({\color{red}{x}}) - 96\]
\[\Large f({\color{red}{1}}) = ({\color{red}{1}})^3 - 15({\color{red}{1}})^2 + 68({\color{red}{1}}) - 96\]
\[\Large f(1) = ???\]
Use PEMDAS to simplify. Tell me what you get.

Answer 15

-43

Answer 16

close but no

Answer 17

Can you explain to me how this one is done

Answer 18

Replace every x with 1 to get
\[\Large f(x) = x^3 - 15x^2 + 68x - 96\]
\[\Large f({\color{red}{x}}) = ({\color{red}{x}})^3 - 15({\color{red}{x}})^2 + 68({\color{red}{x}}) - 96\]
\[\Large f({\color{red}{1}}) = ({\color{red}{1}})^3 - 15({\color{red}{1}})^2 + 68({\color{red}{1}}) - 96\]
\[\Large f(1) = 1 - 15(1) + 68(1) - 96\]
\[\Large f(1) = 1 - 15 + 68 - 96\]
\[\Large f(1) = -14 + 68 - 96\]
\[\Large f(1) = 54 - 96\]
\[\Large f(1) = -42\]

Answer 19

oh wow i was close lol. that makes sense

Answer 20

So whats the next step.

Answer 21

jim can u help me plz im in a hurry

Answer 22

yes you were only off by 1 @Gibby

Answer 23

so x = 1 leads to y = -42

Answer 24

(x,y) = (1,-42) is another ordered pair on this graph

Answer 25

when x = 2, what is the value of y?

Answer 26

What do you mean. Im sorry if I'm being bit complicated. Math is not the subject for me.

Answer 27

plug in x = 2 (i.e., replace every x with 2). What do you get when you simplify?

Answer 28

-20

Answer 29

incorrect

Answer 30

Was i close

Answer 31

kinda

Answer 32

try it again

Answer 33

ok

Answer 34

-12

Answer 35

yes so (2,-12) is another point on this curve

Answer 36

so you just keep doing this to generate multiple points. You will plot the points and then draw a curve through them all. The more points you determine, the more accurate the graph.

Answer 37

Ok. So how many do you suggest i do.

Answer 38

not sure. At least 5 or so points. That way you can see the general shape and direction and such

Answer 39

So when i graph it i just connect the points because doesn't it have to be rounded. i just feel like i wouldn't get it right.

Answer 40

yes you would draw a curve through all the points the best you can. Since it's done by hand, it doesn't have to be perfect

Answer 41

So thats all for this problem?

Answer 42

That's just for the graph. For the x intercepts, you'll have to use a graphing calculator to find them. If you can't use a graphing calculator, then you'll have to make estimates based on your graph

Answer 43

How do i do that by hand?

Answer 44

you mean find the x intercepts? or do the graph?

Answer 45

the x intercepts

Answer 46

you would look to see where the graph crosses or touches the x axis

Answer 47

oh i see ok

Answer 48

since your graph won't be perfect, neither will the x intercept estimates. But that's to be expected

Answer 49

Ok. so is that all i have to do?

Answer 50

yes that will wrap up the problem

Answer 51

Ok thank you! I really appreciate it!

Answer 52

glad to be of help