#1**+2 **

To determine the intercepts of this equation in the form \(ax+by+c=0\), I would just make x and y zero and see what the result is:

**Finding the x-intercept**

For a line without the slope of 0, the line only touches the x-axis once. For a point to be on the x-intercept, y must be zero; otherwise, it would not be on the x-intercept. Knowing this, you can set the y to be zero and solve for x:

\(4x+5y+6=0\) | Make y=0 so that the point is on the x-intercept. |

\(4x+5*0+6=0\) | |

\(4x+6=0\) | Subtract 6 on both sides. |

\(4x=-6\) | Divide by 4 on both sides. |

\(x=\frac{-6}{4}=-\frac{3}{2}=-1.5\) | |

Ok, we have determined that the x-intercept is located exactly on the point \((-\frac{3}{2},0)\)

**Finding the y-intercept**

You will utilize the exact same logic to find the y-intercept. Of course, x will be equal to 0 this time:

\(4x+5y+6=0\) | Substitute 0 in for x. |

\(4*0+5y+6=0\) | |

\(5y+6=0\) | Subtract 6 on both sides. |

\(5y=-6\) | Divide by 5 on both sides. |

\(y=-\frac{6}{5}=-1.2\) | |

Ok, we have determined that the y-intercept is located exactly on the point \((0,-1.2)\).

You actually do not need any more information to graph this equation. Plot both the intercepts on a coordinate plane, and use a ruler to connect them. Then, you are done!

TheXSquaredFactor Jul 28, 2017