How do you graph $y + 4 = \frac{1}{3} (x + 6)$ $y+4=1/3(x+6)$ ?

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Answer 1 · 2018-06-29T14:43:17

See a solution process below:

First, solve the equation for $y$ $y$ to make it easier to work with:

$y + 4 = (\frac{1}{3} \times x) + (\frac{1}{3} \times 6)$ $y + 4 = (1/3 xx x) + (1/3 xx 6)$

$y + 4 = \frac{1}{3} x + 2$ $y + 4 = 1/3x + 2$

$y + 4 - 4 = \frac{1}{3} x + 2 - 4$ $y + 4 - color(red)(4) = 1/3x + 2 - color(red)(4)$

$y + 0 = \frac{1}{3} x - 2$ $y + 0 = 1/3x - 2$

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

Next, solve for two points which solve the equation and plot these points:

First Point: For $x = 0$ $x = 0$

$y = (\frac{1}{3} \times 0) - 2$ $y = (1/3 xx 0) - 2$

$y = 0 - 2$ $y = 0 - 2$

$y = - 2$ $y = -2$ or $(0, - 2)$ $(0, -2)$

Second Point: For $x = 6$ $x = 6$

$y = (\frac{1}{3} \times 6) - 2$ $y = (1/3 xx 6) - 2$

$y = 2 - 2$ $y = 2 - 2$

$y = 0$ $y = 0$ or $(6, 0)$ $(6, 0)$

We can next plot the two points on the coordinate plane:

graph{(x^2+(y+2)^2-0.04)((x-6)^2+y^2-0.04)=0 [-10, 10, -5, 5]}

Now, we can draw a straight line through the two points to graph the line:

graph{(y - (1/3)x + 2)(x^2+(y+2)^2-0.04)((x-6)^2+y^2-0.04)=0 [-10, 10, -5, 5]}

How do you graph y+4=13(x+6)y+4=1/3(x+6)?