Question

PLEASE HELP Find the slope and equation of the tangent to each of the following curves at the given point:

Answer 1

Answer:

Step-by-step explanation:

To find the slope of the tangent at a point we will find the derivative of equation at that point.

a). y = x² - 10x

$\frac{dy}{dx}=\frac{d}{dx}(x^{2} -10x)$

y' = 2x - 10

At (x = 2),

y' = 2(2) - 10

y' = 4 - 10

y' = -6

From the given equation,

y = 2² - 10

y = -6

Therefore, y-coordinate of the point is y = -6

Equation of the tangent at (2, -6) having slope = -6

y - 2 = -6(x - 2)  

y - 2 = -6x + 12

y = -6x + 14

b). y = $x^{2}+\frac{2}{x}$

At x = $\frac{1}{2}$

y = $(\frac{1}{2})^2+\frac{2}{\frac{1}{2}}$

y = $\frac{1}{4}+4$

y = $\frac{17}{4}$

Now we have to find the equation of a tangent at $(\frac{1}{2},\frac{17}{4})$

y' = 2x - $\frac{2}{x^{2} }$

At x = $\frac{1}{2}$

y' = $2(\frac{1}{2})-\frac{2}{(\frac{1}{2})^2}$

y' = 1 - 8

y' = -7

Therefore, equation of the tangent at $(\frac{1}{2},\frac{17}{4})$ will be,

$y-\frac{17}{4}=-7(x-\frac{1}{2})$

y = -7x + $\frac{7}{2}+\frac{17}{4}$

y = -7x + $\frac{31}{4}$