The graph of which function has a minimum located at (4, –3)? f(x) = x2 + 4x – 11 f(x) = –2x2 + 16x – 35 f(x) = x2 – 4x + 5 f(x) = 2x2 – 16x + 35

Step-by-step explanation:

In order to solve the question, we have to derivate each function.

1) f(x) = x2 +4x -11

Then,

f'(x)= 2x +4

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 2*4 +4 = 12 ≠ 0 then this function doesn't not have a minimum at (4, -3)

2) f(x) = –2x2 + 16x – 35

Then,

f'(x)= -4x +16

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= -4*4 16 = 4 ≠ 0 then this function have a critical point at (4, -3)

then,

f''(4) =-4 <0 then we have a minimum at (4, -3)

3) f(x) = x2 – 4x + 5

Then,

f'(x)= 2x -4

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 2*4 -4 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)

4) f(x) =  2x2 – 16x + 35

Then,

f'(x)= 4x -16

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 4*4 -16 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !

step-by-step explanation:

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