x-Intercepts and y-Intercepts: Finding Where a Line Crosses

Set one variable to zero at a time.

$(6, 0)$ and $(0, 6)$

Set $y = 0$: $x = 6$, giving $(6, 0)$.

Set $x = 0$: $y = 6$, giving $(0, 6)$.

$(6, 0)$ and $(0, 8)$

Set $y = 0$: $4x = 24$, so $x = 6$.

Set $x = 0$: $3y = 24$, so $y = 8$.

$(3, 0)$ and $(0, -9)$

The $y$-intercept is the constant, $-9$, giving $(0, -9)$.

Set $y = 0$: $0 = 3x - 9$, so $x = 3$, giving $(3, 0)$.

$(4, 0)$ and $(0, -10)$

Set $y = 0$: $5x = 20$, so $x = 4$.

Set $x = 0$: $-2y = 20$, so $y = -10$.

Try setting $y = 0$ and see what happens.

No.

Setting $y = 0$ gives $0 = 5$, which is false, so no point on the line has height zero.

The line is horizontal at height $5$ and runs parallel to the $x$-axis, never meeting it.

$\left(\tfrac{10}{3}, 0\right)$ and $\left(0, \tfrac{5}{2}\right)$

Set $y = 0$: $3x = 10$, so $x = \tfrac{10}{3}$.

Set $x = 0$: $4y = 10$, so $y = \tfrac{10}{4} = \tfrac{5}{2}$.

The intercepts are two points. Use the slope formula.

$m = \tfrac{2}{5}$

$m = \tfrac{-2 - 0}{0 - 5} = \tfrac{-2}{-5} = \tfrac{2}{5}$.

With the intercept already known, the equation is $y = \tfrac{2}{5}x - 2$.

One intercept is the starting position, the other is when the balance reaches zero.

$(0, -630)$ and $(14, 0)$: they start $630 in debt and clear it after 14 weeks.

Set $x = 0$: $y = -630$, so at the start the balance is $-630$, meaning $630 owed.

Set $y = 0$: $45x = 630$, so $x = 14$.

After $14$ weeks the balance reaches zero and the debt is cleared. Beyond that the value turns positive, which now represents savings rather than debt.