Question

Which set of parametric equations represents the function y=x^2+4x-5? Select all that apply.

Answer 1

Solution

- The way to solve the equation is to take the expression for x i.e. x = 2t, and substitute into the expression for y(x).

- The result must be the corresponding y-value in terms of t.

- This is done below:

Option A:

$\begin{gathered} x=2t \\ y(x)=x^2+4x-5 \\ \\ \text{ put }x(t)=2t \\ \\ y(x(t))=(2t)^2+4(2t)-5 \\ y(x(t))=y(t)=4t^2+8t-5 \\ \\ \therefore y(t)=4t^2+8t-5\text{ \lparen OPTION A\rparen} \end{gathered}$

Option B:

$\begin{gathered} x=t+1 \\ y=x^2+4x-5 \\ \\ y(x(t))=y(t)=(t+1)^2+4(t+1)-5 \\ t^2+2t+1+4t+4-5 \\ y(t)=t^2+6t\text{ \lparen NOT IN THE OPTIONS\rparen} \end{gathered}$

Option C:

$\begin{gathered} x=t-3 \\ y=x^2+4x-5 \\ \\ y(x(t))=(t-3)^2+4(t-3)-5 \\ =t^2-6t+9+4t-12-5 \\ =t^2-2t-8\text{ \lparen NOT IN THE OPTIONS\rparen} \end{gathered}$

Option D:

$\begin{gathered} x=t^2 \\ y=x^2+4x-5 \\ \\ y(x(t))=(t^2)^2+4(t^2)-5 \\ =t^4+4t^2-5\text{ \lparen NOT IN THE OPTIONS\rparen} \end{gathered}$

Option E:

$\begin{gathered} x=t+1 \\ y=x^2+4x-5 \\ \\ y(x(t))=(t+1)^2+4(t+1)-5 \\ =t^2+2t+1+4t+4-5 \\ =t^2+6t\text{ \lparen OPTION E IS CORRECT\rparen} \end{gathered}$

Final Answer

The answers are OPTIONS A AND E