Question

f(x)=(x+8)^2+3
Find the point on the graph of function that is closest to the point (-2, 6).

Answer 1

both the given point, and th epoint on the function would contain the same line, Normal to the curve right

Answer 2

i think you will have to write another function for distance
distance between f(x) and given point
\[\sqrt{(x+2)^2+(f(x)-6)^2}\]

Answer 3

yes, the direction should be normal to the curve,

Answer 4

we can square the expression, because min will occur when the expression under the square root is minimum

Answer 5

yeah minimize the distance

Answer 6

\[d={(x+2)^2+(f(x)-6)^2}\]

Answer 7

just set the derivative of that=0

Answer 8

they give you the function y =

place tha ty value in to the distance function , to get distance in terms of only x

Answer 9

if you doin algebra

Answer 10

f\[f(x)=x^4+32x^3+379x^2+1956x+3725\]

Answer 11

Is that correct?

Answer 12

yea, i guess,
dont call it f(x) you might confuse yourself, call it s(x) or something else

Answer 13

\[d = \sqrt{(x+2)^2+(f(x)-6)^2}\]

from above


replace f(x), for the function value, f(x) =(x+8)2 + 3

Answer 14

simplify all that, so you have distance d in terms of just x,

Answer 15

minimize that function,

Answer 16

you can minimiz d^2 instead of d
\[d^2=x^4+32x^3+379x^2+1956x+3725\]
is what amy is getting

Answer 17

differentiate, equate to zero, solve and find x, and corresponding f(x)

Answer 18

you might need a calculator :P

Answer 19

I want to make sure the function i got is correct first, is it?

Answer 20

or
@DanJS has got me thinking, at the minimum point, the tangent to the curve will be perpendicular to the line connecting point on curve and (-2,6)

Answer 21

yea, ur function looks correct

Answer 22

yeah , use the tangent at that point (x,y) and the normal line will have slope -1/y'

Answer 23

so just take the derivative, f'(x), that is the tangent slope at the point (x,y)

the slope of the line between (-2,6) and (x,y) on the function, has a slope perpendicular to the tangent slope

normal to curve slope = -1/y'

Answer 24

\[s'(x)=-1/4x^3+96x^2+758x+1956\]

Answer 25

by the way , i just put that distance function d(x) into the calc, and found the d ' (x), and solved that for being 0,

x=6

Answer 26

back to the derivative thing


slope of normal line is -1/y ' or -1/(2x + 16

and the components of that slope in x and y directions, forming the right triangle between the points are y-6 and x+2


the slope of that normal -1/(2x+16) = (y - 6) / x + 2)

Answer 27

How did u get -1/4

Answer 28

how did you get 2x+16?

Answer 29

the derivative of the function y ' = 2(x+8)^1*1 = 2x + 16

Answer 30

i thought you're taking the derivative of (x+2)^2+((x+8)^2+3-6)^2

Answer 31

\[s(x)=x^4+32x^3+379x^2+1956x+3725\]
just take the derivative of this;
other stuff was just an alternate method

Answer 32

that is the slope of the tangent at a point on the function

the slope of the line between that point (x,y) and (-2,6) will be perpendicular to that tangent line -1/y ' or -1/(2x+16)

Answer 33

\[s'(x)=-1/4x^3+96x^2+758x+1956\]
this is what you got right?

Answer 34

how did you get the -1/4 at the start?

Answer 35

that should be the same as change in y over change in x. or (y - 6 ) / (x + 2)

Answer 36

\[\large \frac{ -1 }{ 2x+16 } = \frac{ y - 6 }{ x + 2 }\]

replace the function value for the y variable

Answer 37

\[\large \frac{ -1 }{ 2x+16 } = \frac{ [(x+8)^2 + 3] - 6 }{ x + 2 }\]


that solves to x = -6 again, same as the distance formula wau

Answer 38

point on the curve (x,y) = (-6 , f(-6))

Answer 39

so plug -6 into the original function?

Answer 40

i got the answer, thank you for the help

Answer 41

the derivative of that distance, is a cubic function right?, that is hard to solve , more than just doing

-1/y'(x) = (y - y1) / (x - x1)

Answer 42

I understand now. (-6,7)