How to find the equation of the tangent line to

m1cadc

m1cadc

Answered question

2022-02-10

How to find the equation of the tangent line to the curve f(x)=(x33x+1)(x+2) at the point (1, -3)?

Answer & Explanation

Bucharly9

Bucharly9

Beginner2022-02-11Added 11 answers

Explanation:
First, expand the given function so that we can derive it with ease:
f(x)=(x33x+1)(x+2)
f(x)=x4+2x33x25x+2
Then, we differentiate the function to find its gradient/derivative(s):
f(x)=4x3+6x26x5
Since we have been given the coordinates of a point (1, -3) in the curve / tangent line, we know that we need to find the gradient of the tangent line at x=1:
f(1)=413+612615
f'(1)=4+6-6-5
f'(1)=1
Finally, we use the gradient-intercept formula y2y1=m(x2x1) to find the equation of the tangent line:
y- -3=1(x-1)
y+3=x-1
y=x-4

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