Finding the $y$- intercept to a tangent
I've tried to solve this problem, but I am having some problems.
I need to find the $y$- intercept of the tangent to $y = \frac{18}{x^2+2}$ at point $(1,6)$.
I keep getting $0$, but I don't think that's right.
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$\begingroup$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 $y'=-36x/(x^2+2)^2$.$m=y'(1) =-4$ 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.
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