anonymous
  • anonymous
how to find an equation in the form y-k=a(x-h)^2 if the x intercepts are -2 and 6 and the y-intercept is -6?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
if you know the two x-intercepts, then you can add them up and divide by 2 to get the value of h
anonymous
  • anonymous
-2+6 = 4/2 = 2 h=2
jim_thompson5910
  • jim_thompson5910
yep

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
the y intercept is -6 so x = 0 and y = -6 y-k=a(x-h)^2 y-k=a(x-2)^2 ... plug in h = 2 -6-k=a(0-2)^2 ... plug in (x,y) = (0,-6) -6-k = a(-2)^2 -6 - k = 4a -6 - k+6 = 4a+6 -k = 4a + 6 k = -4a - 6 Now let's plug in an x intercept, say (x,y) = (-2,0) y-k=a(x-2)^2 0-k=a(-2-2)^2 -k = a(-4)^2 -k = 16a k = -16a -4a - 6 = -16a ... replace k with -4a-6 now solve for 'a' to get a = ???
anonymous
  • anonymous
a=1/2
jim_thompson5910
  • jim_thompson5910
yes, a = 1/2 k = -4a - 6 k = -4(1/2) - 6 k = ???
anonymous
  • anonymous
k=-8 is the final equation y+8=(1/2)(x-2)^2?
jim_thompson5910
  • jim_thompson5910
yes it is
anonymous
  • anonymous
Oh, thank you very much!
jim_thompson5910
  • jim_thompson5910
you're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.