Question

Question! Screenshot attached! Not really that hard, but I need some help. :)

Answer 1

i think its 1 because that is where the minimum of the equation is if they r calling a the minimum?

Answer 2

Hint:

Vertex form is y = a(x-h)^2+k

if a > 0 (ie if 'a' is positive), then the parabola opens upward
if a < 0 (ie if 'a' is negative), then the parabola opens downward

Answer 3

it's not 1

Answer 4

oh ok i gotcha my bad a is the leading coefficeint that they want in front of the x squared term right?

Answer 5

yes exactly

Answer 6

you are correct

you find this by first finding the vertex (which in this case is (5,1))

Then you use another point to find the value of 'a'

Answer 7

haha its all good :b

Answer 8

i helped on your last one

Answer 9

y = a(x-h)^2 + k

y = a(x-5)^2+1 ... Plug in the vertex

0 = a(6-5)^2+1 ... Plug in another point, say (6,0)

0 = a(1)^2+1

0 = a(1) + 1

0 = a+1

-1 = a

a = -1

Answer 10

any other problems?

Answer 11

y = a(x-h)^2 + k


y = a(x-(-1))^2-4 ... Plug in the vertex (-1,-4)


y = a(x+1)^2-4


0 = a(1+1)^2-4 ... Plug in another point, say (1,0)


0 = a(2)^2 - 4


0 = 4a - 4


Keep going to solve for 'a'

Answer 12

you got it, np