The function given in the question is
[tex]g(x)=-5x-1[/tex]
Question is to find the equation parallel to the line above and that passes through the point
[tex](1,6)[/tex]
The general equation of a line in slope-intercept form is
[tex]y=mx+c[/tex]
Where,
[tex]\begin{gathered} m=\text{slope} \\ c=\text{intercept} \end{gathered}[/tex]
By comparing coefficients, we will have
[tex]m=-5,c=-1[/tex]
For two lines to be parallel, they must have the same value of their slopes
[tex]m_1=m_2=-5[/tex]
The formula used to calculate the equation of a line when the slope and the points are given is
[tex]\begin{gathered} m=\frac{y-y_1}{x-x_1} \\ \text{where,} \\ m=-5,x_1=1,y_1=6 \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} m=\frac{y-y_1}{x-x_1} \\ -5=\frac{y-6}{x-1} \end{gathered}[/tex]
By cross multiplying the equation above, we will have
[tex]\begin{gathered} -5=\frac{y-6}{x-1} \\ y-6=-5(x-1) \\ y-6=-5x+5 \\ \text{collect like terms, we will have} \\ y=-5x+5+6 \\ y=-5x+11 \end{gathered}[/tex]
Therefore,
The final answer is y = -5x + 11
