Question

Find the slope of the curve at the given point P and an equation of the tangent line at P.

y = -3 - x³, (1, -4)

Answer 1

Let the slope of the function $y = -3 - x^3$ be $m_1$

$$m_1 = \frac{dy}{dx}=-3x^2$$

at P(1,-4),

$$m_1 = -3(1)^2 = -3(1)=-3$$

The tangential line to the curve at P(1,-4) is parallel to the curve, thus

$$m_1=m_2$$

where $m_2$ is the slope of the equation of tangential line.

The equation of the tangent line can be obtained thus:

$$m_2 = \frac{y-y_1}{x-x_1}$$

where:

$$x_1=1; y_1=-4$$

$$m_2 = \frac{y-(-4)}{x-1}=\frac{y+4}{x-1}$$

since $m_2=3$ , then

$$3=\frac{y+4}{x-1}\\ 3(x-1)=y+4\\ 3x-3=y+4\\ 3x-y =4+3\\ 3x-y=7 \implies y=3x-7$$

is the required tangential equation.