Question

Clarification: If f(x)= 6x^3=9x^2-x +C and f(-3)=1; C=-245?

Answer 1

I think you have too many equals signs

Answer 2

This is how it is on the paper. I solved it with:

f(x) = 6x^3 + 9x^2-X+C
162+81- (-3) +C
F(-3) =1= 246+C
-246+1=-245
-245=C

Answer 3

6(-3)^3 = -162

Answer 4

??? I don't understand

Answer 5

-162+81+3+c=1
78+c=1
c=-77

Answer 6

That's not an answer choice. I have C=-245; c=79; c=31; and c=-83

Answer 7

sorry -78+c=1
c=79

Answer 8

What did I do wrong? May you explain?

Answer 9

when you cubed the -3, so changed it to positive instead of leaving it as negative

Answer 10

Okay thank you... Can you help me on one more thing please?

Answer 11

I found the Y-intercepts. However for the x-intercepts of f(x)= (x+1) (x-6) (x-1)^2 I'm not sure how to find the x- intercepts.

Answer 12

the x-intercepts are when y=0
(x+1)(x-6)(x-1)(x-1)=0
the zero product rule says if any or all are 0, it is true
so x={-1,1,6}

Answer 13

Okay so i solved it like this: 0= -1(x+1) (x-6) (x-1)^2

Answer 14

why 0=-1(x+1)(x-6)(x-1)^2
why the -1?

Answer 15

Well I did some research and someone did it that way I ended up getting the same answer as you. What did you do?

Answer 16

you did it fine, just when you did 6(-3)^3 you got a positive answer instead of a negative answer. (-3)^3 is -27 not +27

Answer 17

I'm talking about for the intercepts?

Answer 18

the -1 doesn't play any role in finding the intercepts so putting it there wont change the answer but it shouldn't be there

Answer 19

So how would you have solved it?

Answer 20

the x-intercepts are when y=0
(x+1)(x-6)(x-1)(x-1)=0
the zero product rule says if any or all are 0, it is true
so x={-1,1,6}
I did it the same way as you but you put -1(x+1)(x-6)(x-1)(x-1)=0 [ the -1 in front] which doesn't change the answer but shouldn't be there

Answer 21

Okay sorry for all the trouble. Thanks.

Answer 22

Why does -6 turn to 6 then?

Answer 23

(x-6)=0 so x=6
(x+1)=0 so x=-1
(x-1)=0 so x=1

Answer 24

Right! I feel so dumb. Thank you!