Question

find (a) the slope of the curve at the given point p, and (b) an equation of the tangent line at p. y = - 3 - 4x^2; p(5, - 103).

Explanation:

Step1: Differentiate the function

The derivative of $y = - 3-4x^{2}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ and $\frac{d}{dx}(c)=0$ (where $c$ is a constant) is $y^\prime=\frac{d}{dx}(-3)-4\frac{d}{dx}(x^{2})=0 - 4\times2x=-8x$.

Step2: Find the slope at point P

Substitute $x = 5$ into the derivative. When $x = 5$, $y^\prime=-8\times5=-40$. So the slope of the curve at point $P(5,-103)$ is $-40$.

Step3: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,-103)$ is the point on the line and $m=-40$ is the slope.
Substitute these values into the formula: $y-(-103)=-40(x - 5)$.
Simplify the equation: $y + 103=-40x+200$, then $y=-40x + 97$.

Answer:

(a) - 40
(b) $y=-40x + 97$