Find the number x and y if:-a.) (2x-1),y+2)=(1,2)(Specially find x) Ri

1.	Find the number x and y if:-a.) (2x-1),y+2)=(1,2)(Specially find x) Right answer will be thanked will be followed by me and many more.Note:(Don't write if you don't know please it's my humble request please:)
Answer» Answer: Step-by-step explanation: et’s do this one using middle SCHOOL math. We have f(x,y)=0 where f(x,y)=x2–2x+4y2+16y+1 Let’s zoom into a point on the CURVE (r,s) f(x,y)=f(r+(x−r),s+(y−s)) =(r+(x−r))2−2(r+(x−r))+4(s+(y−s))2+16(s+(y−s))+1 =r2+2r(x−r)+(x−r)2−2r−2(x−r))4s2+8s(y−s)+(y−s)2+16s+16(y−s)+1 =r2−2r+4s2+16s+1+(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2 f(x,y)=f(r,s)+(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2 Since (r,s) is assumed on the curve, f(r,s)=0 f(x,y)=(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2 The tangent at (r,s) is the best linear approximation at (r,s) . We simply DROP the second DEGREE terms and set the whole thing to zero: 0=(2r−2)(x−r)+(8s+16)(y−s) For the no slope case (vertical line) the coefficient on y is zero: 8s+16=0 s=−2 0=f(r,−2)=r2–2r+4(−2)2+16(−2)+1 0=r2–2r−15 (r−5)(r+3)=0 r=5,r=−3 Points with vertical tangents: (5,−2) and (−3,−2) plot 0= x^2 – 2x + 4y^2 + 16y + 1 , x=5, x=-3 hope it helps :)

Find the number x and y if:-a.) (2x-1),y+2)=(1,2)(Specially find x) Right answer will be thanked will be followed by me and many more.Note:(Don't write if you don't know please it's my humble request please:)

Answer»

Answer:

Step-by-step explanation:

et’s do this one using middle SCHOOL math. We have f(x,y)=0 where

f(x,y)=x2–2x+4y2+16y+1

Let’s zoom into a point on the CURVE (r,s)

f(x,y)=f(r+(x−r),s+(y−s))

=(r+(x−r))2−2(r+(x−r))+4(s+(y−s))2+16(s+(y−s))+1

=r2+2r(x−r)+(x−r)2−2r−2(x−r))4s2+8s(y−s)+(y−s)2+16s+16(y−s)+1

=r2−2r+4s2+16s+1+(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2

f(x,y)=f(r,s)+(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2

Since (r,s) is assumed on the curve, f(r,s)=0

f(x,y)=(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2

The tangent at (r,s) is the best linear approximation at (r,s) . We simply DROP the second DEGREE terms and set the whole thing to zero:

0=(2r−2)(x−r)+(8s+16)(y−s)

For the no slope case (vertical line) the coefficient on y is zero:

8s+16=0

s=−2

0=f(r,−2)=r2–2r+4(−2)2+16(−2)+1

0=r2–2r−15

(r−5)(r+3)=0

r=5,r=−3

Points with vertical tangents: (5,−2) and (−3,−2)

plot 0= x^2 – 2x + 4y^2 + 16y + 1 , x=5, x=-3

hope it helps