drawing the graph and finding intercepts from equation not in y=mx+c form

Question

I have the equation: $x+4y = 0$.

Its not in the form of $y=mx+c$ so I am unsure of how to find the intercepts and draw the graph. I know that generally when $x+4y=12$ then you just make one variable $= 0$ and find the other variable but since the constant is $0$ I am unsure how to find these variables. So how would I find the points/intercepts of this equation and what would it look like on a graph?

Use exactly the same method you used for $x+4y=12$. Changing the $12$ to $0$ really makes no difference at all. — David, Apr 29 '14 at 06:27
It makes a big difference, because the method of working out the two intercepts and connecting them only works if there are two intercepts. — user11977, Apr 29 '14 at 06:43

score 1 · Answer 1 · answered Apr 29 '14 at 06:30

1

You can put it in y-intercept form. which is $y=mx+b$. The graph would have a negative slope. I think you can go from there and figure it out yourself.

answered Apr 29 '14 at 06:30

score 0 · Answer 2 · answered Apr 29 '14 at 06:28

0

Hint: $x+4y = 0\implies $$\begin{align} y&=-\dfrac14 x+0\\ y&=\ \ \ \ mx+c\\ \end{align}$

answered Apr 29 '14 at 06:28

user103816

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drawing the graph and finding intercepts from equation not in y=mx+c form

2 Answers2