Determine the equation of a tangent at a given point from the equation of a function Calculus

Let f be the function defined as follows :

f:x\longmapsto3x^3-2x

Determine the equation of the tangent to f for x=-1.

y=-8x-3

y=-8x-13

y=-8x+6

y=-8x+3

Let f be the function defined as follows :

f:x\longmapsto\sqrt{2x+7}

Determine the equation of the tangent to f for x=1.

y=\dfrac{1}{6}x+\dfrac{17}{6}

y=\dfrac{1}{3}x+2

y=\dfrac{1}{6}x+2

y=\dfrac{1}{3}x+\dfrac{8}{3}

Let f be the function defined as follows :

f:x\longmapsto x^{2}e^{x}

Determine the equation of the tangent to f for x=-1.

y=-\dfrac{1}{e}x+\dfrac{2}{e}

y=\dfrac{3}{e}x+\dfrac{1}{e}

y=-\dfrac{1}{e}x

y=-ex+1

Let f be the function defined as follows :

f:x\longmapsto \left(2x+5\right)\ln\left(x\right)

Determine the equation of the tangent to f for x=1.

y=7x

y=7x-7

y=7ex+e-7

y=\dfrac{1}{7}x-\dfrac{e}{7}

Let f be the function defined as follows :

f:x\longmapsto x^{4}e^{-3x}

Determine the equation of the tangent to f for x=1.

y=\dfrac{1}{e^{3}}x

y=\dfrac{1}{e^{3}}x+\dfrac{1}{e^{3}}-1

y=\dfrac{2}{e^{3}}x+\dfrac{2}{e^{3}}

y=\dfrac{4}{e^{3}}x+\dfrac{4}{e^{3}}-3

Let f be the function defined as follows :

f:x\longmapsto \dfrac{x^{2}}{x-3}

Determine the equation of the tangent to f for x=4.

y=6x-4

y=-8x+12

y=6x-32

y=-8x+48