help me plz!!!!!!!!!!!!!!!!!!!!

Question

find a point on the curve y=x^3+2x+2

whose tangent line is parallel to the line3x-y =2. is theremore than on e such point? is so find all other points with thisproperty

in steps plzzz

Answer 1

The curve is y = x ³ + 2x + 2

Differentating with respect of x .

y '  = 3x ² + 2

Here y ' is slope of the tangent line and, the tangent line is parallel to 3x - y  = 2.

So parallel lines have same slopes.

Write the equation 3x - y = 2 in slope intercept form y = mx + c , where m is slope and c  is y - intercept.

y =  3x - 2

Slope is  3.

Therefore, slope of the tangent line is also  3.

3x ² + 2 = 3

3x ² = 3 - 2

x ² = 1/3

x  = ± √1/3

x  =  1/√3 and x  = - 1/√3

x  = 0.57 and x  = -0.57

Using the x - value find the corresponding y - value with the curve.

Substitute x = 1/√3 in y = x ³ + 2x + 2 .

y = ( 1/√3)³ + 2(1/√3) + 2

y  = 1/(√3)³  + 2/√3 + 2

y = 0.19 + 1.15 + 2

y = 3.34

Substitute x = - 1/√3 in y = x ³ + 2x + 2 .

y = (-1/√3)³ + 2(-1/√3) + 2

y = - 0.19 - 1.15 + 2

y = 0.66

Therfore the points are (1/√3, 3.34) and (-1/√3, 0.66).