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

Given a function f, the equation of the tangent line to the graph in point x=a is:

y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)

Here we have:

• f\left(x\right)=3x^{2}-2x
• a=-1

Therefore:

y=f'\left(-1\right)\cdot\left(x+1\right)+f\left(-1\right)

Evaluate f\left(-1\right) :

f\left(-1\right)=3\left(-1\right)^{2}-2\left(-1\right)=5

Calculate f'\left(x\right) :

f'\left(x\right)=\left(3x^{2}-2x\right)'=6x-2

Evaluate f'\left(-1\right) :

f'\left(-1\right)=6\left(-1\right)-2=-8

The equation of the tangent line to the graph of the function f at point x=-1 is:

y=-8\left(x+1\right)+5

y=-8x-8+5

Given a function f, the equation of the tangent line to the graph at point x=a is:

y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)

Here, we have:

• f\left(x\right)=\sqrt{2x+7}
• a=1

Therefore:

y=f'\left(1\right)\cdot\left(x-1\right)+f\left(1\right)

Evaluate f\left(1\right) :

f\left(1\right)=\sqrt{2\left(1\right)+7}=\sqrt{9}=3

Calculate f'\left(x\right) :

f'\left(x\right)=\left(\sqrt{2x+7}\right)'=\dfrac{\left(2x+7\right)'}{2\sqrt{2x+7}}=\dfrac{2}{2\sqrt{2x+7}}=\dfrac{1}{\sqrt{2x+7}}

Evaluate f'\left(1\right) :

f'\left(1\right)=\dfrac{1}{\sqrt{2\left(1\right)+7}}=\dfrac{1}{3}

The equation of the tangent line to the graph of the function f at point x=1 is:

y=\dfrac{1}{3}\left(x-1\right)+3

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

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

Given a function f, the equation of the tangent line to the graph at point x=a is:

y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)

Here, we have:

• f\left(x\right)=x^{2}e^{x}
• a=-1

Therefore:

y=f'\left(-1\right)\cdot\left(x+1\right)+f\left(-1\right)

Evaluate f\left(-1\right) :

f\left(-1\right)=\left(-1\right)^{2}e^{-1}=\dfrac{1}{e}

Calculate f'\left(x\right) :

f'\left(x\right)=\left(x^{2}e^{x}\right)'=\left(x^{2}\right)'e^{x}+x^{2}\left(e^{x}\right)'=2xe^{x}+x^{2}e^{x}=\left(2x+x^{2}\right)e^{x}

Evaluate f'\left(-1\right) :

f'\left(-1\right)=\left(-2+1\right) e^{-1}=-\dfrac{1}{e}

The equation of the tangent line to the graph of the function f at point x=-1 is:

y=-\dfrac{1}{e}\left(x+1\right)+\dfrac{1}{e}

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

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

Given a function f, the equation of the tangent line to the graph at point x=a is:

y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)

Here, we have:

• f\left(x\right)=\left(2x+5\right)\ln\left(x\right)
• a=1

Therefore:

y=f'\left(1\right)\cdot\left(x-1\right)+f\left(1\right)

Evaluate f\left(1\right) :

f\left(1\right)=7\ln\left(1\right)=0

Calculate f'\left(x\right) :

f'\left(x\right)=\left(\left(2x+5\right)\ln\left(x\right)\right)'=\left(2x+5\right)'\ln\left(x\right)+\left(2x+5\right)\left(\ln\left(x\right)\right)'=2\ln\left(x\right)+\dfrac{2x+5}{x}

Evaluate f'\left(1\right) :

f'\left(1\right)=2\ln\left(1\right)+7=7

The equation of the tangent line to the graph of the function f at point x=1 is:

y=7\left(x-1\right)+0

y=7x-7

Given a function f, the equation of the tangent line to the graph at point x=a is:

y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)

Here, we have:

• f\left(x\right)=x^{4}e^{-3x}
• a=1

Therefore:

y=f'\left(1\right)\cdot\left(x-1\right)+f\left(1\right)

Evaluate f\left(1\right) :

f\left(1\right)=\left(1\right)^{4}e^{-3\left(1\right)}=e^{-3}=\dfrac{1}{e^{3}}

Calculate f'\left(x\right) :

f'\left(x\right)=\left(x^{4}e^{-3x}\right)'=\left(x^{4}\right)'e^{-3x}+x^{4}\left(e^{-3x}\right)'=4x^{3}e^{-3x}+x^{4}e^{-3x}\left(-3\right)=\left(4x^{3}-3x^{4}\right)e^{-3x}

Evaluate f'\left(1\right) :

f'\left(1\right)=e^{-3}=\dfrac{1}{e^{3}}

The equation of the tangent line to the graph of the function f at point x=1 is:

y=\dfrac{1}{e^{3}}\left(x-1\right)+\dfrac{1}{e^{3}}

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

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

Given a function f, the equation of the tangent line to the graph at point x=a is:

y-f\left(a\right)=f'\left(a\right)\cdot\left(x-a\right)

Here, we have:

• f\left(x\right)=\dfrac{x^{2}}{x-3}
• a=4

Therefore:

y=f'\left(4\right)\cdot\left(x-4\right)+f\left(4\right)

Evaluate f\left(4\right) :

f\left(4\right)=\dfrac{\left(4\right)^{2}}{4-3}=16

Calculate f'\left(x\right) :

f'\left(x\right)=\left(\dfrac{x^{2}}{x-3}\right)'=\dfrac{\left(x^{2}\right)'\left(x-3\right)-x^{2}\left(x-3\right)'}{\left(x-3\right)^{2}}=\dfrac{2x\left(x-3\right)-x^{2}}{\left(x-3\right)^{2}}=\dfrac{x^{2}-6x}{\left(x-3\right)^{2}}

Evaluate f'\left(4\right) :

f'\left(4\right)=\dfrac{\left(4\right)^{2}-6\left(4\right)}{\left(4-3\right)^{2}}=-8

The equation of the tangent line to the graph of the function f at point x=4 is:

y=-8\left(x-4\right)+16

y=-8x+32+16

y=-8x+48.