Question

sketch the curve given by the set of parametric equations.

$x = t^{2}+4$
$y = \frac{t}{5}+8$

Explanation:

Step1: Express \(t\) from \(y\) - equation

From \(y=\frac{t}{5}+8\), we get \(t = 5(y - 8)\)

Step2: Substitute \(t\) into \(x\) - equation

Substitute \(t = 5(y - 8)\) into \(x=t^{2}+4\). Then \(x=(5(y - 8))^{2}+4=25(y - 8)^{2}+4\). This is a parabola that opens to the right (since the coefficient of \((y - k)^{2}\) is positive, where \(k = 8\) and the standard form of a parabola opening to the right is \(x=a(y - k)^{2}+h\) with \(a = 25\) and \(h = 4\))

We can also find some points:
When \(y = 8\), \(t=0\) and \(x = 4\)
When \(y=9\), \(t = 5\) and \(x=25 + 4=29\)
When \(y = 7\), \(t=- 5\) and \(x=25+4 = 29\)

Since the first graph has \(x\) values in the negative range (which is not possible as \(x=t^{2}+4\geqslant4\) for all real \(t\)) and the second graph has a wrong - shaped curve (not a parabola opening to the right), we can conclude that the correct graph is the one that is a parabola opening to the right. But since we are not given the third graph's full details, based on the equation \(x = 25(y - 8)^{2}+4\) (a parabola opening to the right with vertex \((4,8)\)) and domain \(x\geqslant4\) and range \(y\in(-\infty,\infty)\)

Answer:

The graph which is a parabola opening to the right with vertex \((4,8)\) (assuming the third graph in the original multiple - choice (not fully shown here) is the correct one based on the parametric equations' properties)