Answer: x = {-2, 2}
Explanation:
Tangent means it is touching. Find the intersection of the two equations.
Solve the linear equation for y, then set the two equations equal to each other.
![4y=x+8\qquad \rightarrow \qquad y=(x+8)/(4)](https://img.qammunity.org/2021/formulas/mathematics/college/sed43om90zcvzu90fjeotzc03dzbm60mts.png)
![(x-1)/(x)=(x+8)/(4)\\\\\\\text{Cross multiply and solve for x:}\\4(x-1)=x(x+8)\\4x-4=x^2+8x\\.\qquad 0=x^2+4x+4\\.\qquad 0=(x+2)^2\\.\qquad 0=x+2\\.\qquad x=-2](https://img.qammunity.org/2021/formulas/mathematics/college/zrdgv077oe57kohx0dth6zhhfxuyg81ekv.png)
To find the next point that is parallel to the linear equation and tangent to the curve, we need to use the linear equation with slope (m) =
and unknown b.
Let's try b = 0, then the equation of the linear equation is:
![y=(1)/(4)x](https://img.qammunity.org/2021/formulas/mathematics/college/2cu44cl3e70dtwzklnvx58qhpbz20j8exq.png)
Set the equations equal to each other and solve for x:
![(x-1)/(x)=(x)/(4)\\\\\\4(x-1)=x^2\\4x-4=x^2\\.\qquad 0=x^2-4x+4\\.\qquad 0=(x-2)^2\\.\qquad 0=x-2\\.\qquad x=2](https://img.qammunity.org/2021/formulas/mathematics/college/2vch4fj0lcsywmcydrjag65byt6kvvrfgf.png)
This works!!! If it didn't work, we would have tried other values for b until we arrived at a solution.