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Find a point on the curve y=2x^(3)-4x+1 whose tangent?

+1 vote
line is parallel to the line y-2x=1. Is there more than one such point? If so find all other points with this property.
asked Feb 8, 2013 in ALGEBRA 1 by chrisgirl Apprentice

2 Answers

+2 votes

y - 2x = 1.

Add 2x to each side.

y = 2x + 1.

Compare equation with Slope–intercept form y = mc =+ b. and write the coefficients. Slope m = y' = 2.

And y = 2x3 - 4x + 1

Apply derivative each side.

y' = 6x2 - 4 But y' = 2 so, 6x2 - 4 = 2.

Add 4 to each side.

6x2 = 6

Divide each side by 6.

x2 = 1 then x = ±1

If x = 1 then y = 2(1)3 - 4(1) + 1 = 2 - 4 + 1 = -1

If x = -1 then y= 2(-1)3 - 4(-1) + 1 = -2 + 4 + 1 = 3.

Points are (1, -1) and (-1, 3)

Point (1, -1)

Using the slope intercept formula for the tangent line of y=mx+b,

At point (1, -1), the line equation is -1 = 2(1) + b ⇒ b = -3.

The line equation is y = 2x - 3.

Point (-1, 3)

At point (-1, 3), the line equation is 3 = 2(-1) + b ⇒ b = 5.

The line equation is y = 2x + 5.

answered Feb 8, 2013 by richardson Scholar
0 votes

The curve is y = 2x3 - 4x + 1.

Differentiate the curve with respect to ' x '.

y ' = 6x2 - 4  → ( 1 )

This is the slope (m ) of the tangent line to the implicit curve.

From the given data, tangent line is parallel to the line y - 2x = 1.

 

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

Write the equation : y - 2x = 1 in slope - intercept form.

y = 2x + 1.

Compare the above equation with slope - intercept form line equation : y = mx + b.

Slope (m) = 2 → ( 2 )

Because, the parallel lines have same slopes, slope of the tangent line is 2.

From eq (1) & (2)

6x2 - 4 = 2

3x2 - 2 = 1

3x2 = 1 + 2 = 3

x2 = 3/3 = 1

⇒ x = ± 1.

When, x = - 1, y = 2(- 1)3 - 4(- 1) + 1 = - 2 + 4 + 1 = 3.

When, x = 1, y = 2(1)3 - 4(1) + 1 = 2 - 4 + 1 = - 1.

These are the points (- 1, 3) and (1, - 1) on the curve y = 2x3 - 4x + 1.

 

Slope(m) = 2.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (- 1, 3).

3 = (2)(- 1) + b

 b = 3 + 2

b = 5.

The tangent line equation at the point (- 1, 3) is y = 2x + 5.

 

Slope(m) = 2.

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

- 1 = (2)(1) + b

 b = - 1 - 2

b = - 3.

The tangent line equation at the point (1, - 1) is y = 2x - 3.
answered Jul 3, 2014 by lilly Expert

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