#### To determine

**To find:** The equation of the tangent line to the curve at the point.

#### Answer

The equation of the tangent line to the curve
y=(3x−1)−6 at
(0,1) is
y=18x+1.

#### Explanation

**Given:**

The function is
y=(3x−1)−6.

Point is (0,1).

**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.

**Derivative rules:**

**Chain Rule:**

If *h* is differentiable at *x* and *g* is differentiable at
h(x), then the composite function
F=g∘h defined by
F(x)=g(h(x)) is differentiable at *x* and
F′ is given by the product

F′(x)=g′(h(x))⋅h′(x) (2)

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

(2) Power Rule:
ddx(xn)=nxn−1

(3). Difference rule:
ddx(f−g)=ddx(f)−ddx(g)

(4). Constant multiple rule :
ddx(cf)=cddx(f)

**Calculation:**

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

dydx=ddx(y)=ddx((3x−1)−6)

Use chain rule and power rule as,

dydx=(−6)(3x−1)−6−1ddx((3x−1))=−6(3x−1)−7ddx((3x−1))

Use Difference rule and constant multiple rule as,

dydx=−6(3x−1)−7[3ddx(x)−ddx(1)]=−6(3x−1)−7[3(1)−(0)]=−18(3x−1)−7

Therefore, the derivative of *y* is
dydx=−18(3x−1)−7.

The slope of the tangent line at
(0,1) is computed as follows,

m=dydx|x=0=−18(3(0)−1)−7=−18(−1)−7=18

Thus, the slope of the tangent line at
(0,1) is
m=18.

Substitute
(0,1) for
(x1,y1) and
m=18 in equation (1),

(y−1)=18(x−0)y−1=18xy=18x+1

Therefore, the equation of the tangent line to the curve
y=(3x−1)−6 at
(0,1) is
y=18x+1.