Find an equation of the line that is tangent to the graph of F and parallel to the given line.

FUNCTION: f(x)=x^3+2

LINE: 3x-y-4=0



ive tried this a thousand times. i didnt have time at tutor to ask her for help. i need easy step by step instructions please. thanks so much. #SO CONFUSEDDDDD:(

parallel to the given line: means it has same gradient
LINE: 3x-y-4=0
y = 3x - 4
gradient = 3

FUNCTION: f(x)=x^3+2
the line will touch the function at a point where f'(x) = gradient = 3
3 x^2 = 3
x^2 = 1
x1 = 1
x2 = -1

y1 = x^3 + 2 = 1 + 2 = 3
y2 = -1 + 2 = 1

There are 2 point (1, 3) and (-1, 1), which will lead to 2 line (2 answer)

line 1:
y = 3x + b
3 = 3.1 + b
b = 0
y = 3x

line 2:
y = 3x + b
1 = -3 + b
b = 4
y = 3x + 4

answer: y = 3x and y = 3x + 4

3x^2 =3 is setting the derivative of the function, y=x^3 + 2, equal to the derivative of the function to make it tangent. The gradient, a term more used in vector functions, is the derivative.

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I hope this help:

FUNCTION: f(x)=x^3+2 <------ the main function
f '(x) = 3 x^2 + 0 = 3x^2 <--------- derivative of the main function
the line will touch the function at a point where f '(x) = gradient = 3
3 x^2 = 3

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