Question

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

Answer 1

Answer:

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

Answer 2

Answer:

Instantaneous rate of change = -13

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

Step-by-step explanation:

the instantaneous rate of change of the function $f(x) = 6x^2 - x$ at the point (-1,7)

To find instantaneous rate of change we take derivative

$f(x) = 6x^2 - x$

$f'(x) = 12x-1$

Now we plug in  -1 for x

$f'(-1) = 12(-1)-1=-13$

Instantaneous rate of change = -13

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

$f(x) = -x^2 + 5$

To find slope of tangent line take derivative

$f'(x) = -2x$

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

$f'(1) = -2(1)=-2$

so  slope = -2

use equation $y-y1=m(x-x1)$

$y-4=-2(x-1)$

$y-4=-2x+2$

Add 4 on both sides

y= -2x +6