Question

consider the following parametric equations:

$x = \frac{4t - 15}{2}$ and $y = 3 - t$

step 2 of 2: plot the points that result from the integer t-values from zero to five on the graph. round the x - and y - coordinates of each point to one decimal place, if necessary.

Explanation:

Step1: Substitute \(t = 0\)

For \(x=\frac{4t - 15}{2}\), when \(t = 0\), \(x=\frac{4\times0-15}{2}=\frac{- 15}{2}=-7.5\).
For \(y = 3 - t\), when \(t = 0\), \(y=3-0 = 3\). The point is \((-7.5,3)\).

Step2: Substitute \(t = 1\)

For \(x=\frac{4t - 15}{2}\), when \(t = 1\), \(x=\frac{4\times1-15}{2}=\frac{-11}{2}=-5.5\).
For \(y = 3 - t\), when \(t = 1\), \(y=3 - 1=2\). The point is \((-5.5,2)\).

Step3: Substitute \(t = 2\)

For \(x=\frac{4t - 15}{2}\), when \(t = 2\), \(x=\frac{4\times2-15}{2}=\frac{-7}{2}=-3.5\).
For \(y = 3 - t\), when \(t = 2\), \(y=3-2 = 1\). The point is \((-3.5,1)\).

Step4: Substitute \(t = 3\)

For \(x=\frac{4t - 15}{2}\), when \(t = 3\), \(x=\frac{4\times3-15}{2}=\frac{-3}{2}=-1.5\).
For \(y = 3 - t\), when \(t = 3\), \(y=3-3=0\). The point is \((-1.5,0)\).

Step5: Substitute \(t = 4\)

For \(x=\frac{4t - 15}{2}\), when \(t = 4\), \(x=\frac{4\times4-15}{2}=\frac{1}{2}=0.5\).
For \(y = 3 - t\), when \(t = 4\), \(y=3 - 4=-1\). The point is \((0.5,-1)\).

Step6: Substitute \(t = 5\)

For \(x=\frac{4t - 15}{2}\), when \(t = 5\), \(x=\frac{4\times5-15}{2}=\frac{5}{2}=2.5\).
For \(y = 3 - t\), when \(t = 5\), \(y=3-5=-2\). The point is \((2.5,-2)\).

Answer:

The points are \((-7.5,3)\), \((-5.5,2)\), \((-3.5,1)\), \((-1.5,0)\), \((0.5,-1)\), \((2.5,-2)\)