Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1020: 61

Answer

See below:
1571286563

Work Step by Step

We have to graph the provided equations; also, we have to eliminate the common variable t and make a single equation in x and y. The provided parametric equation in y is $ y=2t $. Computing the value of t from this equation, we get: $\begin{align} & y=2t \\ & t=\frac{y}{2} \\ \end{align}$ By putting this value of t in the equation of x, we get: $\begin{align} & x={{t}^{2}}+t+1 \\ & x={{\left( \frac{y}{2} \right)}^{2}}+\frac{y}{2}+1 \\ \end{align}$ And solving this equation further, we get: $\begin{align} & x={{\left( \frac{y}{2} \right)}^{2}}+\frac{y}{2}+1 \\ & x=\frac{{{y}^{2}}}{4}+\frac{y}{2}+1 \\ & x=\frac{{{y}^{2}}+2y+4}{4} \end{align}$ And completing the square, we get: $\begin{align} & x=\frac{{{y}^{2}}+2y+4}{4} \\ & x=\frac{{{y}^{2}}+2y+1+3}{4} \\ & x=\frac{{{\left( y+1 \right)}^{2}}+3}{4} \\ \end{align}$ Then multiplying both sides by 4 and then subtracting 3 from both sides, we get: $\begin{align} & 4x={{\left( y+1 \right)}^{2}}+3 \\ & 4x-3={{\left( y+1 \right)}^{2}} \\ \end{align}$ Then writing the equation in the form of ${{y}^{2}}=4ax $, we get: $\begin{align} & {{\left( y+1 \right)}^{2}}=4x-3 \\ & {{\left( y+1 \right)}^{2}}=4\left( x-\frac{3}{4} \right) \\ \end{align}$ And the above equation is of the form of a parabola, having vertex $\left( \frac{3}{4},-1 \right)$. We see from the equations that the domain and range are: Domain: $[\dfrac{3}{4},\infty]$ Range: $[-\infty,\infty]$
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