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# Math

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151
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+701

http://prntscr.com/lisuzo

Nov 15, 2018

#1
+626
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a) We must let $$f(x)=0$$.

$$3x^3+10x^2-13x-20=0$$, and solving we get $$x=-4, -1, \frac{5}{3}$$ .

Therefore, the x-intercepts are $$x=-4, -1, \frac{5}{3}$$.

b) Let $$x=0$$ , and solving we get $$-20$$ .

Therefore, the y-intercept is $$y=-20$$.

c) This is simple, make a point between $$x=-4$$ and $$x=-1$$, for example, $$(-3, 0)$$.

Same for $$x=-1$$ and $$x=5/3$$ , for example, $$(1, 0)$$ .

d) As $$x$$ goes toward negative infinity, we know that $$x^3$$ will always decrease, and vice versa.

e) Your graph may look something like this:

You are very welcome!

:P

Nov 15, 2018
#2
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!!! :)

Nov 15, 2018
#3
+99736
+1

Let's see how CS  might have determined the x intercepts  (roots)

3x^3 + 10x^2 -13x - 20

By the Rational Roots Theorem ......one possible root is  -1

So

3(-1)^3 + 10(-1)^2 - 13(-1) - 20 =  -3 + 10 + 13 - 20  =  0

Using synthetic division, we can find the remaining polynomial

-1  [  3   10    - 13     -20 ]

-3     - 7       20

__________________

3     7    -20         0

So....the remaining polynomial is   3x^2  + 7x - 20

Factoring, we have

(3x - 5) ( x + 4)

Setting each factor to 0  and solving for x we have that the other two roots are

x= 5/3     and x = -4

Then the x intercepts are  x = -4, -1   and 5/3

The rest of the answer is as CS has shown....!!!

Nov 15, 2018
#4
+701
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thankyouu bothh

Nov 15, 2018
#5
+626
0

You're welcome! ;)

CoolStuffYT  Nov 15, 2018