A curve has equation y=x^3-x+3

Question

what is the equation of tangent and normal to the curve at point (1,3)?

Answer 1

The Equation of the curve is y = x³ -x+3.

Differentiate y with respect to x.

y' = 3x² - 1.

at x = 1, m = 3(1)² - 1

m = 2.

Now the Tangent line Equation is in the form of y = mx + b

This line passes through the point (1, 3)

3 = 2*1 + b

b = 1

Then y = mx+ b

y = 2x + 1.

Then tangent line equation is 2x - y + 1 = 0.

For normal line equation is

We know that slope of perpendicular lines are equal to -1.

m1*m2 = -1

We know that slope of the tangent line is 2 then

2*m2 = -1

m2 = -1/2

Slope of the normal line Equation is -1/2

Then normal line equation y = mx + b.

y = (-1/2)x+b

The normal line equation passes through the point (1 , 3)

3 = -1/2 + b

b = 7/2

Then line equation y = mx+b

y = (-1/2)x + (7/2)

2y = -x + 7

x + 2y - 7 = 0

Then normal line equation is x + 2y - 7 = 0.

Therefore

Tangent line equation is 2x - y + 1 = 0.

Normal line equation is x + 2y - 7 = 0.