What is the slope and intercept for $y + \frac{1}{2} = - \frac{1}{3} (x + \frac{1}{2})$ $y + 1/2 = -1/3(x+1/2)$ and how would you graph it?

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Answer 1 · 2018-06-01T12:10:59

Slope $- \frac{1}{3}$ $-1/3$

Y intercept $(0, - \frac{2}{3})$ $(0,-2/3)$

expand the bracket and subtract $\frac{1}{2}$ $1/2$ from both sides

$y = - \frac{1}{3} x - \frac{1}{6} - \frac{1}{2}$ $y=-1/3x-1/6-1/2$

$- \frac{1}{6} - \frac{1}{2} = - \frac{4}{6} = - \frac{2}{3}$ $-1/6-1/2=-4/6=-2/3$

$y = - \frac{1}{3} x - \frac{2}{3}$ $y=-1/3x-2/3$ this is the standard form of a straight line graph.

The gradient is $- \frac{1}{3}$ $-1/3$ , the y intercept is $- \frac{2}{3}$ $-2/3$

When y=0, x=-2

Put x=-3 into the equation and y= $\frac{1}{3}$ $1/3$

You now have 3 points to plot

$(0, - \frac{2}{3}), (- 2, 0) and (- 3, \frac{1}{3})$ $(0,-2/3), (-2,0) and (-3,1/3)$

What is the slope and intercept for y+12=−13(x+12)y + 1/2 = -1/3(x+1/2) and how would you graph it?