SOLUTION: Find the tangent and normal lines for {{{f(x)=4-x^2}}}

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Question 128195: Find the tangent and normal lines for f%28x%29=4-x%5E2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given function


Derive to get

Now to find the slope of the tangent line at (1,3), simply plug in x=1 into


Plug in x=1


Multiply


So the slope of the tangent line at (1,3) is -2 (note: this is where you made a mistake)





Now let's use the Point-Slope Formula to find the equation of the tangent line


---Point-Slope Formula---
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point

So lets use the Point-Slope Formula to find the equation of the line

y-3=%28-2%29%28x-1%29 Plug in m=-2, x%5B1%5D=1, and y%5B1%5D=3 (these values are given)


y-3=-2x%2B%28-2%29%28-1%29 Distribute -2

y-3=-2x%2B2 Multiply -2 and -1 to get 2

y=-2x%2B2%2B3 Add 3 to both sides to isolate y

y=-2x%2B5 Combine like terms 2 and 3 to get 5



So the equation of the tangent line is y=-2x%2B5



Now simply negate and find the reciprocal of the slope m=-2 to get m=1%2F2 to get the perpendicular slope


Now let's use the Point-Slope Formula to find the equation of the normal line note: the normal line has a slope of m=1%2F2 and goes through (1,3)


---Point-Slope Formula---
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point

So lets use the Point-Slope Formula to find the equation of the line

y-3=%281%2F2%29%28x-1%29 Plug in m=1%2F2, x%5B1%5D=1, and y%5B1%5D=3 (these values are given)


y-3=%281%2F2%29x%2B%281%2F2%29%28-1%29 Distribute 1%2F2

y-3=%281%2F2%29x-1%2F2 Multiply 1%2F2 and -1 to get -1%2F2

y=%281%2F2%29x-1%2F2%2B3 Add 3 to both sides to isolate y

y=%281%2F2%29x%2B5%2F2 Combine like terms -1%2F2 and 3 to get 5%2F2 (note: if you need help with combining fractions, check out this solver)


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Summary:

So the equation of the tangent line is y=-2x%2B5 and the equation of the normal line is y=%281%2F2%29x%2B5%2F2



Notice if we graph y=4-x%5E2, the tangent line y=-2x%2B5, and the normal line y=%281%2F2%29x%2B5%2F2, we get


Graph of y=4-x%5E2 (red), the tangent line y=-2x%2B5 (green), and the normal line y=%281%2F2%29x%2B5%2F2 (blue)



So we can see that y=-2x%2B5 is tangent to y=4-x%5E2 and that y=%281%2F2%29x%2B5%2F2 is perpendicular to y=-2x%2B5. So this visually verifies our answer.