find the slope - intercept equation of the tangent line to the graph of f at the given point. 48)…
find the slope - intercept equation of the tangent line to the graph of f at the given point. 48) f(x)=\frac{6}{x} at (1, 6)
Answer
Explanation:
Step1: Find the derivative of $f(x)$
The function $f(x)=\frac{6}{x}=6x^{-1}$. Using the power - rule $(x^n)' = nx^{n - 1}$, we have $f'(x)=-6x^{-2}=-\frac{6}{x^{2}}$.
Step2: Evaluate the derivative at the given $x$ - value
We want to find the slope of the tangent line at $x = 1$. Substitute $x = 1$ into $f'(x)$. So $m=f'(1)=-\frac{6}{1^{2}}=-6$.
Step3: Use the point - slope form to find the equation of the tangent line
The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,6)$ and $m=-6$. So $y - 6=-6(x - 1)$.
Step4: Rewrite the equation in slope - intercept form
Expand the right - hand side: $y-6=-6x + 6$. Then add 6 to both sides to get $y=-6x+12$.
Answer:
$y=-6x + 12$