Question

Write the equation in slope, intercept form of the line that passes through (-1,11) and is parallel to the graph of y=-8x-2.

Answer 1

The equation in slope-intercept form of the line that passes through
`\((-1, 11)\)` and is parallel to the graph of
`\(y = -8x - 2\)` is
`\(y = -8x + 3\).`

The equation of a line in slope-intercept form is given by
`\(y = mx + b\)`, where \(m\) is the slope and \(b\) is the y-intercept.

For a line parallel to the given line \(y = -8x - 2\), the slopes of the two lines are the same. Therefore, the slope (\(m\)) of the new line is also -8.

Now, we can use the point-slope form of the equation of a line, which is
`\(y - y_1 = m(x - x_1)\),` where
`\((x_1, y_1)\)` is a point on the line and
`\(m\)` is the slope.

Given the point
`\((-1, 11)\)` and the slope
`\(m = -8\)`, substitute these values into the point-slope form:

`\[y - 11 = -8(x - (-1))\]`

Simplify the equation:

`\[y - 11 = -8(x + 1)\]`

Distribute the -8:

`\[y - 11 = -8x - 8\]`

Now, isolate \(y\) by adding 11 to both sides of the equation:

`\[y = -8x - 8 + 11\]`

Combine the constants:

`\[y = -8x + 3\]`

So, the equation in slope-intercept form of the line that passes through
`\((-1, 11)\)` and is parallel to the graph of \
`(y = -8x - 2\` ) is
`\(y = -8x + 3\).`