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Juliette [100K]
3 years ago
9

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

Mathematics
1 answer:
Harrizon [31]3 years ago
4 0

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}

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