Answer:
[tex]\displaystyle \large{y = - \frac{3}{2}x}[/tex]
Step-by-step explanation:
Hey there! To find a tangent line of curve at specific point, we have to differentiate a given function. This is because a tangent line has same slope as the specific point and slope is equivalent to dy/dx.
From a given function, we can write as [tex]\displaystyle \large{f(x) = 6x^{-1}}[/tex] via law of exponent where [tex]\displaystyle \large{a^{-n}=\frac{1}{a^n}}[/tex] .
Then differentiate with respect to x using power rules.
Power Rules
For every differentiatable continuous functions, if [tex]\displaystyle \large{f(u) = u^n \to f\prime (u) = nu^{n-1} \cdot u\prime}[/tex] for u is any function and n is all real number.
Therefore:-
[tex]\displaystyle \large{f(x) = 6x^{-1}}\\\displaystyle \large{f\prime(x) = (-1)(6)x^{-1-1} \cdot x\prime}\\\displaystyle \large{f\prime(x) = -6x^{-2} }[/tex]
Therefore, our derived function is [tex]\displaystyle \large{f\prime(x) =- 6x^{-2}}[/tex]
Then substitute x = -2 in a derived function or slope.
[tex]\displaystyle \large{f\prime(-2) = -6x^{-2}}\\\displaystyle \large{f\prime(-2) = -6(-2)^{-2}}\\\displaystyle \large{f\prime(-2) = -6 \cdot \frac{1}{(-2)^2}}\\\displaystyle \large{f\prime(-2) = -\frac{6}{4}}\\\displaystyle \large{f\prime(-2) = -\frac{3}{2}}[/tex]
Therefore, the slope at x = -2 is -3/2.
Now we have to use point-slope formula to find the equation of tangent line.
Point-Slope
[tex]\displaystyle \large{y-y_1 = m(x-x_1) \to y-f(x_1) = f\prime (x_1)(x-x_1)}[/tex]
We may have to find the (x,f(x)) coordinate first. We know x = -2 but we don’t know f(x) coordinate at x = -2 yet. We substitute x = -2 in f(x) to find f(-2).
[tex]\displaystyle \large{f(-2) = \frac{6}{-2}}\\\displaystyle \large{f(-2) = -3}[/tex]
Therefore, our coordinate is (-2,-3).
Substitute in point-slope form.
[tex]\displaystyle \large{y-(-3)=-\frac{3}{2}(x-(-2))}\\\displaystyle \large{y+3=-\frac{3}{2}(x+2)}[/tex]
Then convert to slope-intercept form by isolating y-term and simplify the expression.
[tex]\displaystyle \large{y = - \frac{3}{2}(x+2)+3}\\\displaystyle \large{y = - \frac{3}{2}x-3+3}\\\displaystyle \large{y = - \frac{3}{2}x}[/tex]
Therefore, the equation of tangent line is y = -3x/2.
