Question

Equation of the tangent at (2,-3) when the function is x^2 + y^2 = 13 ??

Answer 1

Find the derivative
2x+2ydy/dx=0
dy/dx=-x/y
m at (2,-3)=2/3
Now use y+3=(2/3)(x-2)

Answer 2

y' = -x/y

Answer 3

can't i use the point given (2,-3) and sub that into the y=mx+b for when m=2/3 ??

Answer 4

hellllllllllllloooooooooooooooo ?

Answer 5

since we found the slope at any point on the function by taking the dirivative to be -x/y... you plug in x and y and get a slope of 2/3 because two neg. equal a positive. then you use point slop to find the actual linear equation of the tangent line at that point.

Answer 6

right I get that part.. but when you plug in the slope you have y=2/3x+b and you need to plug in a point (x,y) to find b. so i'm asking if you plug in the same point given in the question (2,-3) to find b ?

Answer 7

yea, use "point slope formula withat (x1,y1) = (2,-3) and slope = 2/3
point slope formula:
y-y1=m(x-x1) you know m,y1, and x1...

Answer 8

why can't it be (-3)=(2/3)(2) + b ?

Answer 9

so b= -13/3

Answer 10

y-(-3)=(2/3)(x-2) is the equation...that you need to solve in the end...

Answer 11

what's the equation in terms of x?

Answer 12

i dont know what you mean in terms of x ?

Answer 13

b does = -(13/3) good job! walk through the whole thing a few more times to get the hang...