# What is k if f(x)=x^2-kx-3 and f(2)=9 ?

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Given f(2) = 9 , we have to find k in the function f(x) = x^2 - kx - 3.

Now f(x) = x^2 - kx - 3, for x =2

=> f(2) = 2^2 - k*2 - 3 = 9

=> 4 - 2k - 3 = 9

=> 2k = 4 - 3 - 9

=> 2k = -8

=> k = -8/2 = -4

**Therefore the value of k is -4 and the function f(x) is x^2 + 4x - 3.**

We'll substitute x by 2 in the expression of f(x):

f(2) = 2^2-k*2-3

f(2) = 4 - 2k - 3

We'll combine like terms and we'll get:

f(2) = 1 - 2k

From enunciation, we know that f(2) = 9:

1 - 2k = 9

We'll isolate -2k to the left side. For this reason, we'll subtract 1 both sides:

-2k = 9 - 1

-2k = 8

We'll divide by -2:

k = 8/-2

**k = -4**

f(x) = x^2-kx-3. f(2) = 9.

Since f(2) = 9,

f(2) = 2^2-k(2)-3 = 9.

Therefore 4 -2k -3 = 9.

-2k+4-3 = 9.

-2k +1 = 9

Subtract 1 from borh sides and we get:

-2k = = 9-1

-2k = 8

Divide both sides by -2 and we get:

k = 8/-2 = -4.

Therefore k = -2.

We rewrite f(x) using k= -4.

f(x) = x^2 - (-4)x -3 .

f(x) = x^2+4x -3.