Elementary Algebra

To find the \(y\)-intercept, we substitute \(\alert{0}\) for \(x\) in the equation and solve for \(y\text{.}\)

The \(y\)-intercept is \((0,4)\text{,}\) as we saw above. To find the \(x\)-intercept, we substitute \(\alert{0}\) for \(y\) in the equation and solve for \(x\text{.}\)

The \(x\)-intercept is \((-8,0)\text{,}\) as expected.

First, we find the \(x\)- and \(y\)-intercepts of the graph. To find the \(y\)-intercept of the graph, we substitute \(0\) for \(x\) and solve for \(y\text{.}\)

The \(y\)-intercept is the point \((0,3\dfrac{1}{2})\text{.}\)

To find the \(x\)-intercept of the graph, we substitute \(0\) for \(y\) and solve for \(x\text{.}\)

The \(x\)-intercept is the point \((2\dfrac{1}{3}, 0)\text{.}\)

Here is a table showing the two intercepts. We plot the intercepts and connect them with a straight line to obtain the graph below.

\(x\)	\(y\)
\(0\)	\(3\dfrac{1}{2}\)
\(2\dfrac{1}{3} \)	\(0\)

It is a good idea to find a third point as a check. We choose \(x=\alert{1}\) and solve for \(y\text{.}\)

You can check that the point \((1,2)\) lies on the graph.

1. An equation for the temperature is

   \begin{equation*} T=-12+2h \end{equation*}

   A graph of the equation is shown below.

   

2. To find the \(T\)-intercept, we set \(h=0\) and solve for\(T\text{.}\)

   \begin{equation*} T=-12 + 2(\alert{0})=-12 \end{equation*}

   The \(T\)-intercept is \((0,-12)\text{.}\) This point tells us that when \(h=0,~ T=-12\text{,}\) or the temperature at noon was \(-12 \degree\text{.}\)

   To find the \(h\)-intercept, we set \(T=0\) and solve for \(h\text{.}\)

   \begin{align*} \alert{0} \amp = -12+2h \amp \amp \blert{\text{Add 12 to both sides.}}\\ 12 \amp = 2h \amp \amp \blert{\text{Divide both sides by 2.}}\\ 6 \amp = h \end{align*}

   The \(h\)-intercept is the point \((6,0)\text{.}\) This point tells us that when \(h=6,~T=0\text{,}\) or the temperature will reach zero degrees at six hours after noon, or 6 pm.