Question

the circle (x+1)^2+(y-4)^2=26 intersects y axis at?

Answer 1

The graph will intersect the y-axis when x=0. Substituting x=0 into the equation of the circle leaves you with
\((0+1)^2+(y-4)^2=26 \implies 1+(y-4)^2=26 \implies (y-4)^2=25\).

Can you take things from here and solve for y? :-)

Answer 2

ok will try

Answer 3

1 is (0.9) whats the other @ChristopherToni

Answer 4

can u answe r @ChristopherToni

Answer 5

\[(y-4)^2=25 \\ \text{ gives two equations }\]

Answer 6

\[y-4= \pm \sqrt{25}\]

Answer 7

then

Answer 8

thanks i got that

Answer 9

you already solved one of them
so solve the one you didn't solve yet

Answer 10

ty:)

Answer 11

np

Answer 12

the answer is (9,0) and (-1,0) right?

Answer 13

@myininaya

Answer 14

y-4=5 or y-4=-5
y=9 or y=-1
well almost
you let x=0
so the points are (0,9) and (0,-1)

Answer 15

yeh sorry its (0,9) and (0,-1)

Answer 16

anyways good job :)

Answer 17

ty:)