Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.1 Graph Quadratic Functions in Standard Form - Guided Practice for Example 3 - Page 238: 5

Answer

See the graph

Work Step by Step

Given: $y=2x^2+6x+3$ The coefficients are $a =2$, $b =6$, and $c=3$. Because $a > 0$, the parabola opens up. Find the vertex. $x=-\frac{b}{2a}=\frac{-6}{2.2}=-\frac{3}{2}$ Then find the y-coordinate of the vertex. $y=2(-\frac{3}{2})^2+6(-\frac{3}{2})+3=-\frac{3}{2}$ Draw the axis of symmetry $x =-\frac{3}{2}$ The y-intercept is $3$. Plot the point $(0, 3)$. Then reflect this point in the axis of symmetry to plot another point, $(-3,3)$. Evaluate the function for another value of $x$, such as $x =-4$. $y=2(-4)^2+6(-4)+3=11$ Plot the point $(-4, 11)$ and its reflection $(1, 11)$ in the axis of symmetry. Draw a parabola through the plotted points.
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