Question

find x intercepts of y=2(x-1)^2-8. I keep getting +/-3 but that's wrong, why?

Answer 1

4*2 - 8

x-1 = 4

x-1 = -4

Answer 2

ugh ... typoed lol

Answer 3

x-1 = 2 or -2

Answer 4

I get -1 and 3

Answer 5

-3-1 = -4

-4^2 = 16

16-8 not= 0

3-1 = 2

Answer 6

my answer says x int. are (3, 0) and (-1, 0). I'm adding the 8, dividing by 2 and square rooting to get x-1=+/-3 but when i plug it in to graph the x intercepts are -1 and 3

Answer 7

your not doing the last part correctly

Answer 8

\[2(x-1)^2-8=0\]
\[2(x-1)^2=8\]
\[(x-1)^2=4\]
\[(x-1)=\pm\sqrt{4}=\pm2\]
\[x=1\pm2\]

thus \[x=-1,3\]

Answer 9

ahh, got it, so I was just combining them incorrectly

Answer 10

That's great zarkon,I just plotted the graph and got the same result.

Answer 11

Thanks for pointing out my mistake zarkom! (I tend to get the overall concept but then make stupid little errors that confuse me.. :) )

Answer 12

So, to find the x intercept set y = 0 in the equation and then $$ 2 (x - 1)^2 - 8 = 2 (-3+x) (1+x)=0\Rightarrow x=-1,3$$

Answer 13

\[2(x-1)^2-8\]
\[2[(x-1)^2-4]\]
\[2[(x-1-2)(x-1+2)]\]...