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]
