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

PLEASE HELP Find The Slope And Equation Of The Tangent To Each Of The Following Curves At The Given Point class=

Sagot :

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

[tex]\frac{dy}{dx}=\frac{d}{dx}(x^{2} -10x)[/tex]

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 = [tex]x^{2}+\frac{2}{x}[/tex]

At x = [tex]\frac{1}{2}[/tex]

y = [tex](\frac{1}{2})^2+\frac{2}{\frac{1}{2}}[/tex]

y = [tex]\frac{1}{4}+4[/tex]

y = [tex]\frac{17}{4}[/tex]

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

y' = 2x - [tex]\frac{2}{x^{2} }[/tex]

At x = [tex]\frac{1}{2}[/tex]

y' = [tex]2(\frac{1}{2})-\frac{2}{(\frac{1}{2})^2}[/tex]

y' = 1 - 8

y' = -7

Therefore, equation of the tangent at [tex](\frac{1}{2},\frac{17}{4})[/tex] will be,

[tex]y-\frac{17}{4}=-7(x-\frac{1}{2})[/tex]

y = -7x + [tex]\frac{7}{2}+\frac{17}{4}[/tex]

y = -7x + [tex]\frac{31}{4}[/tex]