m1cadc

2022-02-11

What is the equation of the tangent line of $f\left(x\right)=6x-{x}^{2}$ at x=-1?

stefjumnmt

Beginner2022-02-12Added 14 answers

Explanation:

We are given

$f\left(x\right)=6x-{x}^{2}$

To find the equation of the tangent line, we need to: find the slope of the tangent line, obtain a point on the line, and write the tangent line equation.

To find the slope of the tangent line, we take the derivative of our function.

f'(x)=6-2x

Substituting our point x=-1

f'(-1)=6-2(-1)=6+2=8

Now that we have our slope, we need to find a point on the line. We have an x-coordinate, but we need a f(x) too.

$f(-1)=6(-1)-{(-1)}^{2}=-6-1=-7$

So the point on the line is (-1, -7).

With a slope and a point on the line, we can solve for the equation of the line.

$y-{y}_{p}=m(x-{x}_{p})$

y-(-7)=8(x-(-1))

y+7=8x+8

y=8x+1

Hence, the tangent line equation is: f(x)=8x+1

We are given

To find the equation of the tangent line, we need to: find the slope of the tangent line, obtain a point on the line, and write the tangent line equation.

To find the slope of the tangent line, we take the derivative of our function.

f'(x)=6-2x

Substituting our point x=-1

f'(-1)=6-2(-1)=6+2=8

Now that we have our slope, we need to find a point on the line. We have an x-coordinate, but we need a f(x) too.

So the point on the line is (-1, -7).

With a slope and a point on the line, we can solve for the equation of the line.

y-(-7)=8(x-(-1))

y+7=8x+8

y=8x+1

Hence, the tangent line equation is: f(x)=8x+1

regresavo552

Beginner2022-02-13Added 14 answers

Explanation:

we require the slope m and a point (x,y) on the line

and