#### To determine

**To find:** The equation of two lines that are tangent to the curve and are parallel to given line.

#### Answer

The equations of two tangent lines are y=3x−7 and y=3x−3.

#### Explanation

**Given:**

The curve is y=x3−3x2+3x−3 and the line is 3x−y=15.

**Derivative rules:**

(1) Derivative of constant function: ddx(c)=0

(2) Power Rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(4) Difference rule: ddx(f−g)=ddx(f)−ddx(g)

(5) Constant multiple rule: ddx(cf)=cddx(f)

**Formula Used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

Obtain the slope of the tangent line to the curve.

The derivative of curve *y* is dydx, which is obtained as follows,

dydx=ddx(x3−3x2+3x−3)

Apply the sum rule (3) and the difference rule (4),

dydx=ddx(x3−3x2)+ddx(3x−3)=ddx(x3)−ddx(3x2)+ddx(3x)−ddx(3)

Apply the derivative rule (1) and the constant multiple rule (5),

dydx=ddx(x3)−3ddx(x2)+3ddx(x)−ddx(3)=ddx(x3)−3ddx(x2)+3ddx(x)−0

Apply the power rule (2) and simplify the expression,

dydx=(3x3−1)−3(2x2−1)+3(1x1−1)=(3x2)−3(2x1)+3(1x0)=3x2−6x+3

The derivative of the curve *y* is 3x2−6x+3.

Therefore, slope of the tangent of the curve y=x3−3x2+3x−3 is 3x2−6x+3.

Obtain the tangent points.

The given line can be rewritten as, y=3x−15 and its slope is 3.

Since the required equation of the tangent lines is parallel to the given line, the slope of the tangent line is equal to the slope of the given equation.

That is, the slope of the equation 3x2−6x+3 is 3.

Solve the equation 3x2−6x+3=3 and obtain the value of *x* as follows,

3x2−6x=03x(x−2)=0x=0,x=2

Substitute 2 for *x* in y=x3−3x2+3x−3 and obtain the value of *y*.

y=x3−3x2+3x−3=23−3(22)+3(2)−3=8−3(4)+6−3=−1

Thus, the value of y=−1.

Substitute 0 for *x* in y=x3−3x2+3x−3 and obtain the value of *y*.

y=x3−3x2+3x−3=0−0+0−3=−3

Thus, the value of y=−3.

Therefore, the tangent points are (2,−1) and (0,−3).

Obtain the equation of the tangent lines to the curve.

Substitute (2,−1) for (x1,y1) and 3 for *m* in equation (1),

y+1=3(x−2)y+1=3x−6y=3x−6−1y=3x−7

Thus, the equation of tangent line is y=3x−7_.

Substitute (0,−3) for (x1,y1) and 3 for *m* in equation (1).

y+3=3(x−0)y+3=3xy=3x−3

Thus, the equation of the tangent line is y=3x−3_.

Therefore, the equation of the tangent lines to the curve that is parallel to the line 3x−y=15 are y=3x−7 and y=3x−3.