100 points and brainliest answer what is the instantaneous rate of change of the function f(x) = 6x^2 - x at the point (-1,7) a. -13 b. -5 c. 7 d. 11 what is the equation of the line tangent to the function f(x) = -x^2 + 5 at the point (1,4) a. y = -x + 5 b. y = -x + 9 c. y = -2x + 6 d. y = -2x + 9

Just took the test,  

1)d, 3

2)a, -13

3) c, y=-2x+6

4) c, f'(a)=8a+2

Step-by-step explanation:

For #3, graph f(x)=-x^2+5. Next graph y=-2x+6. You will see that the added equation touches at exactly the point (1,4). this make it the tangent line equation

Instantaneous rate of change = -13

Equation of tangent line : y= -2x +6

Step-by-step explanation:

the instantaneous rate of change of the function at the point (-1,7)

To find instantaneous rate of change we take derivative

Now we plug in  -1 for x

Instantaneous rate of change = -13

the equation of the line tangent to the function f(x) at the point (1,4)

To find slope of tangent line take derivative

Plug in 1 for x . given point is (1,4)

so  slope = -2

use equation

Add 4 on both sides

y= -2x +6

my guess is

step-by-step explanation: