Need to find the y -intercept of the tangent to y = 18 x

Step 1: Compute the tangent line.
- The tangent line is a line passing through the point $(1,6)$ with the same slope as the curve that that point.
- In order to write down a line, you need a point on the line and the slope of the line. You already have a point, but you need to find the slope of the line.
- The slope of the line is the derivative at the point $(1,6)$, since the function is
$\frac{18}{x^2+2},$
its derivative is
$-\frac{36x}{(x^2+2{)}^2}$
plugging in $x=1$ from the point $(1,6)$, we get that the slope is
$-\frac{36}{3^2}=-4.$
This is the slope of the line of interest.
- Therefore, using the point-slope form of a line, you get that the tangent line is $y-6=(-4)(x-1)$.
Step 2: Find the $y$-intercept.
- You have the point-slope form of a line, to get $y-6=-4(x-1)$. We can turn this into slope-intercept form to get $y=-4x+10$.
- Therefore, your $y$-intercept is $10$.

First you find the slope of tangent line at the point of tangency. Slope of tangent line is derivative of the function evaluated at the point of tangency. For $y=18/(x^2+2)$ we use quotient rule to get $y^{\prime }=-36x/(x^2+2{)}^2$. $m=y^{\prime }(1)=-4$ is the slope and $y=6-4(x-1)$ is the equation of the tangent line. To find the y intercept we let $x=0$ and get $y=10.$ Thus the point $(0,10)$ is the $y$ intercept of the tangent line.