## anonymous one year ago what is the focus of the parabola given by the equation y=x^2-2x-3? a. (4,-15/4) b. (1,4) c. (1,-4) d. (1,15/4) e. (1,-15/4)

1. misty1212

i swear i just saw this question a second ago

2. misty1212

trick is to write it as $y+3=x^2-2x$ then complete the square on the left do you know how to do that?

3. misty1212

lol i mean on the RIGHT

4. anonymous

no i dont :( @misty1212

5. misty1212

ok ready? it is not hard

6. anonymous

7. misty1212

what is half of 2?

8. misty1212

...

9. anonymous

1 @misty1212

10. misty1212

ok and $$1^2=1$$ so we go right from $y+3=x^2-2x$ to $y+3+1=(x-1)^2$ or $y+4=(x-1)^2$

11. misty1212

now we need the focus, but first we need the vertex the vertex we see from our eyeballs is $$(1,-4)$$

12. misty1212

then we compare to the standard form $4p(y-k)=(x-h)^2$ and see that in this case $$4p=1$$so $$p=\frac{1}{4}$$

13. misty1212

so we travel $$\frac{1}{4}$$ units up from $$(1,-4)$$ and get to $$(1,-\frac{15}{4})$$

14. misty1212

what to check it

15. anonymous