Two points only is not enough to establish a quadratic, unless one of those points is the vertex.
If one of the points is a vertex, use the method I just showed you in your previous post.
Let me know what you think.
how do you determine the the quadratic function if the problem gives you a graph with the line has points of (-3,5) and (0,-4)?
6 answers
the (-3,5) is the vertex how do i use that i am not completely understand
where do i plug in thoes number
Look back at my explanation in the previous post
for vertex (h,k), we get
y = a(x-h)^2 + k
so for vertex (-3,5), we get
y = a(x-(-3) )^2 + 5
or
y = a(x + 3)^2 + 5
Now plug in the other point (0, -4)
-4 = a(0+3)^2 + 5
-4 = a(9) + 5
-9 = 9a
a = -1
y = -(x+3)^2 + 5
for vertex (h,k), we get
y = a(x-h)^2 + k
so for vertex (-3,5), we get
y = a(x-(-3) )^2 + 5
or
y = a(x + 3)^2 + 5
Now plug in the other point (0, -4)
-4 = a(0+3)^2 + 5
-4 = a(9) + 5
-9 = 9a
a = -1
y = -(x+3)^2 + 5
how did you get the nine after the A?
mmmhhh?
(0+3)^2
= 3^2
= 9
(0+3)^2
= 3^2
= 9