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

Need to find the $y$-intercept of the tangent to $y=\frac{18}{{x}^{2}+2}$ at point $\left(1,6\right)$.
I keep getting $0$, but I don't think that's right.
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Corinne Choi
Step 1: Compute the tangent line.
- The tangent line is a line passing through the point $\left(1,6\right)$ 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 $\left(1,6\right)$, since the function is
$\frac{18}{{x}^{2}+2},$
its derivative is
$-\frac{36x}{\left({x}^{2}+2{\right)}^{2}}$
plugging in $x=1$ from the point $\left(1,6\right)$, 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 $y-6=\left(-4\right)\left(x-1\right)$.
Step 2: Find the $y$-intercept.
- You have the point-slope form of a line, to get $y-6=-4\left(x-1\right)$. We can turn this into slope-intercept form to get $y=-4x+10$.
- Therefore, your $y$-intercept is $10$.
###### Not exactly what you’re looking for?
studovnaem4z6
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 $y=18/\left({x}^{2}+2\right)$ we use quotient rule to get ${y}^{\prime }=-36x/\left({x}^{2}+2{\right)}^{2}$. $m={y}^{\prime }\left(1\right)=-4$ is the slope and $y=6-4\left(x-1\right)$ is the equation of the tangent line. To find the y intercept we let $x=0$ and get $y=10.$ Thus the point $\left(0,10\right)$ is the $y$ intercept of the tangent line.