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)
𝑓(𝑥) = 𝑥2 − 4𝑥+1 at point (0,1)
𝑓(𝑥) = 𝑥3+ 2 at point (1,3)
𝑓(𝑥) = 2𝑥4+ 3𝑥3 − 2𝑥 + 7 at point (-1,8)
𝑓(𝑥) = √2𝑥 + 7 𝑎𝑡 𝑥 =1
1
Expert's answer
2022-05-17T11:15:43-0400
1.
f′(x)=(x2−4x+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)=(x3+2)′=3x2
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)=(2x4+3x3−2x+7)′=8x3+9x2−2
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)=(2x+7)′=2x+71
slope=m=f′(1)=2(1)+71=31
f(1)=2(1)+7=3
y−3=31(x−1)
The standard (slope-intercept form) equation of the tangent line is
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