Question

Want an expert to verify if I got the correct solutions!Someone who is good w/ Trig

Answer 1

`y=2x^2+8x+12;[-5,-1]`

Step-by-step explanation

We want to find the rectangular form of the given parametric equation:

`y(t)=2t^2+4`

on the interval [-3, 1] where:

`x(t)=t-2`

To do this, first, write t in terms of x:

`\begin{gathered} x=t-2 \\ t=x+2 \end{gathered}`

Now, substitute the expression for t into the given equation and simplify:

`\begin{gathered} y(x)=2(x+2)^2+4 \\ y(x)=2(x^2+4x+4)+4 \\ y(x)=2x^2+8x+8+4 \\ y(x)=2x^2+8x+12 \end{gathered}`

That is the rectangular form.

To find the interval (x1, x2), we have to substitute the values of the boundaries of the interval (t1, t2) = [-3, 1] into the equation for x:

`\begin{gathered} x_1=t_1-2=-3-2=-5 \\ x_2=t_2-2=1-2=-1 \end{gathered}`

Hence, the interval is:

`[-5,-1]`

That is the answer.