2022-02-11

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

stefjumnmt

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.
$f\left(-1\right)=6\left(-1\right)-{\left(-1\right)}^{2}=-6-1=-7$
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.
$y-{y}_{p}=m\left(x-{x}_{p}\right)$
y-(-7)=8(x-(-1))
y+7=8x+8
y=8x+1
Hence, the tangent line equation is: f(x)=8x+1

regresavo552

Explanation:
we require the slope m and a point (x,y) on the line
${m}_{tangent}={f}^{\prime }\left(-1\right)$
$⇒{f}^{\prime }\left(x\right)=6-2x$
$⇒{f}^{\prime }\left(-1\right)=6+2=8$
and $f\left(-1\right)=-6-1=-7⇒\left(-1,-7\right)$
$⇒y+7=8\left(x+1\right)$
$⇒y=8x+1←$ equation of tangent

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