Answer to Question #341746 in Calculus for Ally

Question

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

Answer 1

1.

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

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

$$y-1=-4(x-0)$$

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

$$y=-4x+1$$

2.

$$f'(x)=(x^3+2)'=3x^2$$

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

$$y-3=3(x-1)$$

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

$$y=3x$$

3.

$$f'(x)=(2x^4+ 3x^3 − 2𝑥 + 7)'=8x^3+9x^2-2$$

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

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

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

$$y=-x+7$$

4.

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

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

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

$$y-3=\dfrac{1}{3}(x-1)$$

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

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