can someone help me with a problem please!!

- heyitslizzy13

can someone help me with a problem please!!

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- heyitslizzy13

##### 1 Attachment

- heyitslizzy13

@satellite73 @Directrix @jim_thompson5910

- Jhannybean

which part of the problem?

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## More answers

- heyitslizzy13

The first part for now :)

- freckles

Hint: Intermediate Value theorem

- heyitslizzy13

@freckles do you know any good websites to review the theorem?

- Jhannybean

\[f(x) = ax^3-11x^2+bx+c \qquad f(-3)=10, \qquad f(0)=4,\qquad f(-1)=4\]\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10 \]\[f(-3)=-27a-99-3b+c=10\]\[f(-3) = -27a-3b+c=109\] ...perhaps we can solve this as a system of equations?

- heyitslizzy13

could you show me how? please and thank you :)

- jim_thompson5910

part a)
imagine you have a river that is a complete straight line
|dw:1443850668196:dw|
this straight line would stretch forever in both directions

- jim_thompson5910

now let's say we had point A and point B
one point on either side
|dw:1443850712927:dw|

- jim_thompson5910

if we connect the two points with a continuous function, then it is impossible to not pass through the line.
we might have something like this
|dw:1443850761686:dw|
the path has to cross over the river somewhere

- heyitslizzy13

okay

- jim_thompson5910

now imagine the river is the x axis
if point A has a positive y coordinate (ie it's above the x axis) and point B has a negative y coordinate (ie it's below the x axis)
then the continuous function must cross the x axis at some point between A and B. So if we know A is above the x axis and B is below, there is at least one root in between

- Jhannybean

\[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c=4\]\[f(-1) =-a-b+c=15\] \[\therefore \]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\]
As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be \(c=4\) if i'm not mistaken.

- heyitslizzy13

I am kind of confused, how do we know that c=4 is the root that lies in the middle?

- Jhannybean

because when \(x=0\) (that means it's at the origin) \(y=4\)

- heyitslizzy13

ohhh okay

- jim_thompson5910

Jhannybean plugged in x = 0 and everything went away basically except for that last term

- Jhannybean

Exactly

- heyitslizzy13

do you understand what part a of the problem is asking? I'm quite confused

- Jhannybean

Okay so between \(x=-3\) and \(x=0\), is there ANYWHERE on the x-axis when the function gives you a solid number? hint: look at the above posts and our explanations.

- jim_thompson5910

if you need a visual way to approach this, draw a blank xy axis
|dw:1443851396111:dw|

- heyitslizzy13

at -1 i believe , (-1, 4) ?

- jim_thompson5910

then plot the point (-3,10), from the fact that f(-3) = 10
and the point (0,4), because f(0) = 4
|dw:1443851450292:dw|

- jim_thompson5910

now, ask yourself this: is it possible to connect the two points without crossing the x axis?

- heyitslizzy13

it is possible

- jim_thompson5910

so there is no guarantee we have a root on the interval from x = -3 to x = 0

- jim_thompson5910

now IF we had (0,-4) instead, then we shift down the second point
|dw:1443851620081:dw|

- jim_thompson5910

|dw:1443851638079:dw|

- jim_thompson5910

now it's impossible to connect the two points without crossing the x axis
so there has to be at least one root between the two x values
|dw:1443851682979:dw|
or
|dw:1443851692524:dw|
or
|dw:1443851699567:dw|

- heyitslizzy13

ohhh okay!

- jim_thompson5910

so really the only guarantee that we have a root is if there is a sign change (in my example the sign change goes from y = 10 to y = -4)

- jim_thompson5910

sign change in the y values or function outputs I mean

- jim_thompson5910

but since y = 10 to y = 4 has no sign changes in the problem, there is no guarantee there's a root between x = -3 to x = 0

- heyitslizzy13

oh i see now

- heyitslizzy13

could you help me on part b?

- jim_thompson5910

ok so I think you found that c = 4 right?

- jim_thompson5910

do you see how Jhannybean got this?
\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\]

- heyitslizzy13

yes :)

- heyitslizzy13

wait i don't understand how she got the 109, 4 and15

- jim_thompson5910

I think f(-3) should have been 10 and not 109

- heyitslizzy13

okay, and how is f(-1)= 15?

- jim_thompson5910

it should be
\[f(-3) = -27a-3b+c=10\\ f(0) =c=4 \\f(-1)=-a-b+c=4\]

- heyitslizzy13

oh okay, i understand then

- jim_thompson5910

oh wait, there's an 11 in there that I missed, let me rethink

- jim_thompson5910

Jhannybean has it right, I missed a term

- Jhannybean

\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10\]\[f(-3) = -27a -99 -3b +c = 10\]\[f(-3) = -27a -99 -3b + c +99 = 10 + 99\]\[f(-3) = -27a -3b+c=109\]

- Jhannybean

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean
\[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = 0+0+0+c=4 \\ f(0)=c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c+11=4+11\]\[f(-1) =-a-b+c=15\] \[\therefore\]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\]
As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be \(c=4\) if i'm not mistaken.
\(\color{#0cbb34}{\text{End of Quote}}\)

- heyitslizzy13

oh ok

- Jhannybean

Does that make more sense, @heyitslizzy13 ?

- heyitslizzy13

yes :)

- Jhannybean

Great \(\checkmark\)

- jim_thompson5910

`that root would be c=4 ` I don't know about that part

- heyitslizzy13

so we know c is 4, what about a and b?

- Jhannybean

Haha, I meant \(x=0\) x_x

- jim_thompson5910

ah I gotcha

- heyitslizzy13

would it be 109 and 15?

- jim_thompson5910

since c = 4 we can replace each 'c' with 4
\[-27a-3b+\color{red}{c}=109\\-a-b+\color{red}{c}=15\]
\[-27a-3b+\color{red}{4}=109\\-a-b+\color{red}{4}=15\]

- Jhannybean

So then you would solve it either by substitution, or using the elimination method :)

- heyitslizzy13

can you walk me through either one please? I know what they are but I'm a little lost rn

- Jhannybean

First we want to isolate a and b in `both` equations to one side, therefore we need to start by subtracting -4 to both sides of both the equations. \[-27a-3b+4-4=109 -4\]\[-a-b+4-4=15-4\] Simplifying both of these, we get: \[-27a-3b=105\]\[-a-b=11\] With me so far?

- heyitslizzy13

yes!

- Jhannybean

So I personally like using the elimination method. How that works is we're trying to eliminate one of the variables by using a multiplier so when we add the two equations together, we'll be left with one variable, then we can use that to solve for the other. Here, i'm going to eliminate \(b\).
in order to eliminate \(b\), i need to multiply the second equation by \(-3\). \[-27a-3b=105\]\[\color{red}{-3}(-a-b=11)\] \[-27a-3b=105\]\[\color{red}{+3}a\color{red}{+3}b=\color{red}{-33}\] Adding these two equations together, we will get \[(-27+3)a +(-3+3)b =105-33\]\[-24a+0b=72\]Now we just solve for \(a\).

- Jhannybean

Is this making sense, @heyitslizzy13 ?

- heyitslizzy13

yes!! :)) is a=-3?

- Jhannybean

Yes it is :)

- Jhannybean

Now we just plug \(a=-3\) back into one of the two equations and we can solve for \(b\) I'm going to plug it into \(-a-b=11\) . \[-(-3)-b=11\]\[3-b=11\]\[-3+3-b=11-3\]\[-b=8\]\[b=-8\]

- heyitslizzy13

okay thanks :)))

- heyitslizzy13

can you help me on c also?

- jim_thompson5910

for part c) I would graph the function with a graphing calculator
if you don't have one, use something like desmos
https://www.desmos.com/calculator

- heyitslizzy13

thanks!!

- jim_thompson5910

the cool thing about desmos, as compared to a TI83, is that you can use sliders to change the graph
https://www.desmos.com/calculator/tvrsgalven

- heyitslizzy13

thank you both so much!!!! :)))

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