Wednesday

July 23, 2014

July 23, 2014

Posted by **kris** on Monday, August 28, 2006 at 1:43pm.

how do I do this?

When x=3, one wants y to be 10

y= (x+a)^3 + 2

20= (3+a)^3 + 2

solve for a. I would change it to

18=(3+a)^3 and take the cube root of each side to solve for a.

I think we need to make it

10=(x+a)^3 + 2 or

8=(x+a)^3 so 8^(1/3)=x+a, then solve this when x=3

Then substitute that into y=(x+a)^3 + 2.

thanks for your help but I am still confused of how to find a right like i got up to here:8=(3+a)^3

then I have to cube it but how? like what would i end up with? I am a little confused!

Ok, here's what we know.

y=x^3 + 2

We first determine the value of x that corresponds to y=10, thus

10=x^3 + 2, which means 8=x^3 and x=2

We know that we want the value x=2 to now be shifted to the right to correspond to x=3. This means we need to subtact 1 from x in shifting to the right (it seems a little counter-intuitive, but study it a bit.)

Thus y=(x-1)^3+2 is the desired shift.

Substitute x=3 and we have y=(3-1)^3+2=2^3 + 2=10.

What would be the shift if we wanted the same graph to coincide with (3,2)?

How about (-4,10)?

Take the cube root of each side...

2= a+3 solve for a. My solution above is wrong, a typo.

repost if you have questions.

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