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)

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.