Question

Complete the square to change the following equation into vertex form:

y^2 + 12y + x^2 - 4x = 41

Answer 1

Ok, @mathmale let me see if I've learned enough from you, @xForeverASaintx do you have a formula or a step by step example in your textbook of one of these problems?

Answer 2

@doulikepiecauseidont No I don't know how to do it.

Answer 3

we can make it into a circle.

\(\large\color{blue}{ y^2 + 12y + x^2 - 4x = 41 }\)
\(\large\color{blue}{ y^2 + 12y \color{red} { +36 } + x^2 - 4x \color{green} { +4 }= 41\color{red} { +36 } \color{green} { +4 } }\)
\(\large\color{blue}{ (y+6)^2+ (x-2)^2= 81 }\)
\(\large\color{blue}{ (y+6)^2+ (x-2)^2= 9^2 }\)

\(\large\color{blue}{ (y-\color{red} { -6 })^2+ (x-\color{red} { 2 })^2=\color{red} { 9 }^2 }\)

Answer 4

@SolomonZelman Is that the vertex form?

Answer 5

you have a squared "y" and "x" that means, is NOT a parabola, and thus it doesn't really have a vertex, or U-turn point

squared "y" and "x" usually means hyperbola or ellipse or circle

in this case both are squared, thus is an ellipse, with major and minor axis equal, thus a circle

so it doesn't have a vertex form per se

Answer 6

but as SolomonZelman showed above, that's the standard form of a circle

Answer 7

y^2 + 12y + x^2 - 4x = 41: I agree with JD: There's no such thing as a "vertex form" for a circle. I think you're referring to finding the CENTER of this circle.

Answer 8

I think it is an unsuccessful attempt of her online lesson to ask her not only to label the center and the etc., of the shape, but also the what the shape is.

Answer 9

@xForeverASaintx: where do you stand now in terms of understanding this problem? need any further help?

Answer 10

it is quite obvious to say that I am adding to numbers to form some perfect squares.

Answer 11

@xForeverASaintx: Still interested, or not? Would appreciate knowing.