👤

Answer :

Solution:

Given:

[tex]\begin{gathered} Coordinates: \\ (5-t,t^2+4t-1) \end{gathered}[/tex]

This implies that:

[tex]\begin{gathered} x=5-t \\ y=t^2+4t-1 \end{gathered}[/tex]

Using the coordinate x,

[tex]\begin{gathered} x=5-t \\ Making\text{ t the subject of the formula;} \\ t=5-x \end{gathered}[/tex]

Hence, substituting t into coordinate y,

[tex]\begin{gathered} y=t^2+4t-1 \\ y=(5-x)^2+4(5-x)-1 \\ Expanding\text{ the brackets;} \\ y=25-10x+x^2+20-4x-1 \\ Collecting\text{ the like terms together and simplfying;} \\ y=25+20-1-10x-4x+x^2 \\ y=44-14x+x^2 \\ \\ Rearranging\text{ the terms:} \\ y=x^2-14x+44 \end{gathered}[/tex]

Therefore,

[tex]y=x^2-14x+44[/tex]