Put the slope you found into $y = mx + b$ and use one point from the table to solve for $b$, the $y$-intercept of h.

Translating a line straight down changes only its $y$-values (and $y$-intercept), not its slope. After updating the equation for this vertical shift, set $y=0$ and solve for $x$ to get the x-intercept.

In an expression line, type (160-130)/(23-18) and note the value Desmos gives; this is the slope of line h.

In a new expression line, type 130 - (slope)*18, replacing slope with the value from Step 1 (for example, 130 - 6*18); the result is the $b$ value in $y = mx + b$ for line h.

Enter the equation for line h as y = 6x + 22. Then, to represent the line translated down 5 units, enter y = 6x + 17 for line k. You should see two parallel lines, with line k below line h.

Add another expression y = 0 to show the x-axis. Click or tap the intersection point of y = 6x + 17 and y = 0; read off the x-coordinate of this intersection. That x-value is the x-coordinate of the x-intercept of line k.

Because h is a line, we can use any two points from the table to find its slope $m$.

Using $(18,130)$ and $(23,160)$:

So line h has slope $m=6$.

A line with slope $m$ has the form $y = mx + b$.

Substitute $m=6$ and one of the points on h, say $(18,130)$:

Since $6\cdot18 = 108$, we get $130 = 108 + b$, so $b = 22$.

Thus line h has equation $y = 6x + 22$.

Translating a graph down 5 units subtracts 5 from every $y$-value, which changes an equation $y = mx + b$ to $y = mx + (b - 5)$; the slope stays the same.

Here $b = 22$, so for line k the new $y$-intercept is $22 - 5 = 17$.

Therefore line k has equation $y = 6x + 17$.

The x-intercept is where the line crosses the x-axis, so $y = 0$.

For line k, solve $0 = 6x + 17$.

Subtract 17 from both sides: $-17 = 6x$, so $x = -\dfrac{17}{6}$.

Therefore, the x-intercept of line k is $\left(-\dfrac{17}{6},0\right)$.