Question

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

Answer 1

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

`(dy)/(dx)=(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)+(2)/(x)`

At x =
`(1)/(2)`

y =
`((1)/(2))^2+(2)/((1)/(2))`

y =
`(1)/(4)+4`

y =
`(17)/(4)`

Now we have to find the equation of a tangent at
`((1)/(2),(17)/(4))`

y' = 2x -
`(2)/(x^(2) )`

At x =
`(1)/(2)`

y' =
`2((1)/(2))-(2)/(((1)/(2))^2)`

y' = 1 - 8

y' = -7

Therefore, equation of the tangent at
`((1)/(2),(17)/(4))` will be,

`y-(17)/(4)=-7(x-(1)/(2))`

y = -7x +
`(7)/(2)+(17)/(4)`

y = -7x +
`(31)/(4)`