# Find the equationof the tangent and the normal to the curve y=2x^3-x+5 at the point (1,6)?

asked Feb 27, 2014 in CALCULUS

Point slope form is y-y1 = m(x-x1)

y = m(x-x1)+y1 ---> (1)

Where m is slope, (x1,y1) = (1,6) are point on the line.

Now find f'(x)

f(1) = 2-1+5 = 6

f(6) = 432-6+5 = 431

f'(x) = f(b)-f(a)/b-a

= (431-6)/(6-1)

= 425/5

= 85

For a line tangent to a curve

Slope m = f'(x), (x1,y1) = (1,6)

From (1)

y = 85(x-1)+6

y = 85x-85+6

y = 85x-79

For a line normal to a curve is

Slope is m = -1/f'(a) , (x1,y1) = (1,6)

From (1)

y = -1/85 (x-1)+6

y = -x/85+1/85+6

y = -x/85 +511/85

y = -x/85+6.01

The tangent line equation to the curve y =  2x3 - x + 5 at the point (1, 6).is y = 5x + 1.

The normal line equation to the curve y =  2x3 - x + 5 at the point (1, 6) is y = (- 1/5)x + (31/5).

The curve y =  2x3 - x + 5 and the point is (1, 6).

y ' = 6x2 - 1.

When, x = 1, y ' = 6(1)2 - 1 = 6 - 1 = 5.

y ' = 5.

This is the slope (m ) of the tangent line to the implicit curve at (1, 6).

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = 5x  + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (1, 6).

6 = 5(1) + b

b = 6 - 5

⇒ b = 1.

The tangent line equation  is y = 5x + 1.

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (1, 6) is - 1/5.

Slope (m) = - 1/5.

Now, the normal line equation is y = (- 1/5)x + b.

Find the y - intercept by substituting the the point in the normal line equation say (x, y) = (1, 6).

6 = (- 1/5)(1) + b

b = 6 + 1/5

b = (30 + 1)/5

b = 31/5.

The normal line equation  is y = (- 1/5)x + (31/5).