Alright faggots, I need help finishing my math homework. I'm supposed to find the intercepts of 9x^2 + 4y = 36

Alright faggots, I need help finishing my math homework. I'm supposed to find the intercepts of 9x^2 + 4y = 36.

Here are my answers: X: (0,-2) (0,2)
Y: (-9,0), (9,0)

Normally I would have factored by moving the 36 to the other side of the equation, to get 9x^2 + 4y - 36 = 0. But when you try to factor it into two binomials by multiplying the -36 and the 9, you're left with -324. As far as I can tell, there are no two factors of -324 that add up to 4. So instead, I just reworked the equation into (9x^2 -36)/4 = y, and (√4y + 36)/9 = x. From there it's pretty easy to just figure out what values would give you zero, so that's how I wound up with my points. I'm just worried that maybe that route won't work because this equation isn't technically in quadratic form. Of course my textbook doesn't have the fucking answer in it, so idk. What does Sup Forums think?

You dont need to factor anything. Just set x=0 to find the y-intercepts, and y=0 to find the x-intercepts

fucking just do

9x^2-36=4y

(9/4)(x^2-4) = y

(9/4)(x-2)(x+2)=y

set y = 0 for x=-2,2

and set x=0 for y=-9 there is no positive value

Aren't you just supposed to put 0 in to the formule for x if you want to find the y intercept and in y if you want the x intercept and just derive the whole thing you get?

\thread

>getting two y intercepts
did you even think about what a plot of this graph would look like?

First line should have -4y on the right

how the fuck does a parabola intercept the y-axis twice?

Use wolfram alpha to get an idea of what youre working woth. Its a parabola, not a circle. If it was y^2 then youd get two y ints

Guy probably doesn't know how to recognize the shape of a graph just from its general formula. He's studying intercepts fer chrissake.

9x^2 + 4y = 36
4y = 36 - 9x^2
Y = 36/4 - (9/4)x^2
Y = 9 - (9/4)x^2
Let x = 0
Y = 9 - (9/4)*0^2
Y = 9 (hits the y axis)
Let y = 0
9 - (9/4)x^2 = 0
X^2 = 9/(9/4)
X^2 = 4
X = 2 (hits the x axis)
X = -2 (hits the x axis)
Therefore point one when x = 2
Y = 9-(9/4)(2)^2
Y = 9 - 4
Y = 5
Thus (2,5)
When x = -2
Y = 5 (because -2^2 is always positive)
Thus (-2,5) is another point
The graph will look like an upside down parabola passing through the x axis at 2 and -2 and the y axis at 9
These are ur axis intercepts, i forgot what the other 2 points that i wrote where useful for, maybe for well i lost the train of thought. Anyway hope that helped :)

When I was sin highschool I would spend hours in a graphing program just playing around with graphs. I was the only one in the class who had any idea what any of the graphs should look like

Circle

x-int:
let y=0
9x^2=36
x^2=4
x=-2 or x=2

y-int:
let x=0
4y=36
y=9

ez pz.
stay in school op

You must be over 18 to post on this site

A circle could intercept the y axis twice, but this isn't a circle.

But you wanted to know what could intercept the y axis twice

He hasn't explicitly stated his age, and i only report people dumb enough to do that.

No you could have a diagonal parabola.

>circle
>y isn't even squared
are you retarded?

I am not him, and that is not what he asked. In fact, you can have a parabola that intersects Y twice. It would just be a formula where there is a Y^2 somewhere in there.

Are you in 8th grade or is burger education that bad

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