Question

find the point on the graph of f(x)=x^2 that is closest to (2 1/2)

Answer 1

2 1/2 ?

Answer 2

2,1/2

Answer 3

@amistre64?

Answer 4

define -1/y': i get -1/(2x)

for any given point (a,b), the slope of the line is -1/(2a)

you want this to define the line: y = -1/(2a) (x-2)+1/2

or theres a different approach using a circle centered at (2,1.2)

Answer 5

(x-2)^2 + (y-1/2)^2 - r^2 = x^2 - y

Answer 6

\[(x-2)^2 + (y-1/2)^2 - r^2 = x^2 - y\]
\[x^2-4x+4 + y^2-y - r^2 = x^2 - y\]
\[-4x+4 + y^2 - r^2 = 0\]
\[r^2=y^2-4x+4\]

Answer 7

ok lets focus on the calculus based one

Answer 8

which is the first one you showed me right?

Answer 9

yes

Answer 10

ok so I understand the first step, your simply finding the derivative

Answer 11

The next step I get as well which is the slope

Answer 12

finding the derivate, and peroping the slope yes

f(x)=x^2
f'(x) = 2x ; -1/f' = -1/(2x)

Answer 13

but then you lost me at the third step though