Graph the line with slope 5 and y-intercept -3

asked by guest
on Mar 28, 2025 at 8:19 am



MathBot Answer:

Slope-Intercept Form:

y=mx+b\begin{aligned} y = mx+b\end{aligned}y=5x+b3=5×0+b3=0+bb=3\begin{aligned}y &= 5x + b \\ -3 &= 5 \times 0 + b \\ -3 &= 0 + b \\ b &= -3 \end{aligned}An equation of the line in slope-intercept form is: y=5x3y = 5 x - 3

Point-Slope Form:

yy1=m(xx1)\begin{aligned} y - y_{1} = m(x - x_{1}) \end{aligned}where m =5, x1=0, and y1=3 \text{where m } = 5, \text{ } x_{1} = 0, \text{ and } y_{1} = -3 An equation of the line in point-slope form is: y+3=5xy + 3 = 5 x

Standard Form:

Ax+By+C=0\begin{aligned}Ax + By + C = 0\end{aligned}Slope=ABAB=51A=5,B=1\begin{aligned} \text{Slope} &= -\frac{A}{B} \\ -\frac{A}{B} &= \frac{5}{1} \\ A = -5, B = 1 \end{aligned}5x+1y+C=05(0)+1(3)+C=00+3+C=00+C=3C=3\begin{aligned}-5 x + 1 y + C &= 0 \\ -5(0) + 1(-3) + C &= 0 \\ 0 + -3 + C &= 0 \\ 0 + C &= 3 \\ C &= 3 \end{aligned}An equation of the line in standard form is: 5xy3=05 x - y - 3 = 0
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