A quadratic function has its vertex at the point (-9,8). The function passes through the point (9,4). When written in vertex form, the function is f(x)= a(x-h)^2 +k, where

a)
h)-9
k)8

4=a(9)^2+8
4=a(9+9)^2+8
4=a(18)^2+8
4=a(332)
a=0.01204

I was able to find both h & k. but the answer I got for a is incorrect, please help me

from vertex,

y = a(x+9)^2 + 8
but (9,4) lies on it, so

4 = a(18^2) + 8
324a = -4
a = -4/324 = -1/81
etc

you messed up here:
4=a(18)^2+8
4=a(332) , you added 8 to 324a , you can't add unlike terms

To solve for the value of a, we can use the quadratic function and the given points (-9, 8) and (9, 4).

We know that the vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) represents the vertex. In this case, the vertex is (-9, 8). Therefore, we can rewrite the equation as:

f(x) = a(x + 9)^2 + 8

Since the function passes through the point (9, 4), we can substitute these coordinates into the equation and solve for a:

4 = a(9 + 9)^2 + 8
4 = a(18)^2 + 8
4 = 324a + 8
324a = -4
a = -4/324
a = -1/81

Therefore, when written in vertex form, the quadratic function is:

f(x) = (-1/81)(x + 9)^2 + 8

So, the correct value for a is -1/81.