Answer to Question #341746 in Calculus for Ally

Question #341746

Equation of a Tangent Line: Find the standard (slope-intercept form) equation of the tangent line to the following functions at the specified points. (Show your complete solution, use Method 1)

  1. 𝑓(𝑥) = 𝑥2 − 4𝑥+1 at point (0,1)
  2. 𝑓(𝑥) = 𝑥3+ 2 at point (1,3)
  3. 𝑓(𝑥) = 2𝑥4+ 3𝑥3 − 2𝑥 + 7 at point (-1,8)
  4. 𝑓(𝑥) = √2𝑥 + 7 𝑎𝑡 𝑥 =1
1
Expert's answer
2022-05-17T11:15:43-0400

1.


f(x)=(x24x+1)=2x4f'(x)=(x^2-4x+1)'=2x-4

slope=m=f(0)=2(0)4=4slope=m=f'(0)=2(0)-4=-4

y1=4(x0)y-1=-4(x-0)

The standard (slope-intercept form) equation of the tangent line is


y=4x+1y=-4x+1

2.


f(x)=(x3+2)=3x2f'(x)=(x^3+2)'=3x^2

slope=m=f(1)=3(1)2=3slope=m=f'(1)=3(1)^2=3

y3=3(x1)y-3=3(x-1)

The standard (slope-intercept form) equation of the tangent line is


y=3xy=3x



3.


f(x)=(2x4+3x32𝑥+7)=8x3+9x22f'(x)=(2x^4+ 3x^3 − 2𝑥 + 7)'=8x^3+9x^2-2

slope=m=f(1)=8(1)3+9(1)22=1slope=m=f'(-1)=8(-1)^3+9(-1)^2-2=-1

y8=(x(1))y-8=-(x-(-1))

The standard (slope-intercept form) equation of the tangent line is


y=x+7y=-x+7

4.


f(x)=(2x+7)=12x+7f'(x)=(\sqrt{2x+7})'=\dfrac{1}{\sqrt{2x+7}}

slope=m=f(1)=12(1)+7=13slope=m=f'(1)=\dfrac{1}{\sqrt{2(1)+7}}=\dfrac{1}{3}

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

y3=13(x1)y-3=\dfrac{1}{3}(x-1)

The standard (slope-intercept form) equation of the tangent line is


y=13x+83y=\dfrac{1}{3}x+\dfrac{8}{3}


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