Question

find (a) the slope of the curve at the given point p, and (b) an equation of the tangent line at p. y = x^2 + 5; p(5,30) (a) the slope of the curve at p(5,30) is . (type an integer or a decimal.)

Explanation:

Step1: Differentiate the function

The derivative of $y = x^{2}+5$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $y'=\frac{d}{dx}(x^{2}+5)=\frac{d}{dx}(x^{2})+\frac{d}{dx}(5)=2x+0 = 2x$.

Step2: Find the slope at the given point

Substitute $x = 5$ into the derivative $y'$. When $x = 5$, $y'(5)=2\times5=10$. This is the slope of the curve at point $P(5,30)$.

Step3: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,30)$ and $m = 10$.
So $y-30 = 10(x - 5)$.
Expand the right - hand side: $y-30=10x-50$.
Add 30 to both sides to get the slope - intercept form: $y=10x - 20$.

Answer:

(a) 10
(b) $y = 10x-20$