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Sketch the plane curve represented by the given parametric equations. Then use interval notation to give the relation's domain and range. x = 2t, y = t^2 + t + 3
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x = 2t, so t = x/2. Substitute this in y=t^2+t+3: y = (x/2)² + x/2 +3\[y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+3\]You don't need the parameter t. The graph of this formula is a parabola. Its vertex is at -b/2a, which is (-1/2 )/(1/2) = -1. The y=value there is 1/4 -1/2 + 3 = 2.75, which is the smallest value of y. You can guess from here what the range is...
Will D.Domain: (-∞, ∞); Range: 2.75, ∞
That's it!
Thank You, I was stumped on that one.
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