Question

pt. on graph of f(x)=x^2-1 closest to (-4,0)?
i'm getting non-real #'s when solving d/dx for 0

Answer 1

let me know if you are still here

Answer 2

yes, on 2nd effort with a little help I got (√5/2,3/2)?

Answer 3

best method to use is lagrange method

Answer 4

i used distance formula plugging in pts. and f(x)=y=(x^2-1), then derive and set to zero?

Answer 5

F= (x+4)^2 + y^2

\[\nabla F=<2(x+4),2y>\]

\[\nabla F=<-2x,1>\]

Answer 6

second one should have been
\[\nabla G=<-2x,1>\]

Answer 7

\[2(x+4)=-\lambda 2x\]
\[2y=\lambda\]
y-x^2=1

Answer 8

gotta solve for x and y

Answer 9

do my distance formula/derive/minimize results look correct?

Answer 10

I think so

Answer 11

mahalo

Answer 12

I got (0,-1)

Answer 13

distance of 4.12

Answer 14

(0,-1) is the vertex, no? concave up parabola?

Answer 15

you numbers
\[\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{\sqrt{5}}{2}+4\right)^2}\]
=5.33

Answer 16

y=x^2 -1

(0,-1) is on the curve

y=0^2 -1
y=-1

it would be concave up