Find the equation of the tangent line to f(x)=-3x^2+x-5 at x=-1

The equation of the tangent line to given f(x)  in the explicit (slope) form is

y= 7x-2

Step-by-step explanation:

First we must find y coordinate of the point in which tangent touch the function

f(-1)= -3(-1)∧2 +(-1)-5 = -3-1-5= -9  A(-1,-9)=(xa,ya)

We know the equation of the tangent

y - ya = f'(xa) (x - xa)

Now we will find f'(x)

f'(x)= (-3x∧2+x-5)' = (-3x∧2)' + (x)' - (5)' = -6x+1-0 = -6x+1

Now we will find f'(xa) = f' (-1) = -6* (-1) + 1 = 6 + 1 = 7

f'(-1) = 7

now we will replace this value in the equation of the tangent and get

y - (-9) = 7 (x-(-1)) => y+9 = 7(x+1) => y+9 = 7x + 7 => y = 7x+7-9

y = 7x-2

Good luck!!!

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step-by-step explanation: