Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 103: 65

Answer

$(a)$ $x$-intercept $A(3,0)$ $y$-intercept $B(0,-6)$ $(b)$ $x$-intercept $B(-1,0)$ $y$-intercept $A(0,-1)$ Also see the images below.

Work Step by Step

$(a)$ To graph the given linear function, we need at least $2$ points corresponding this function. But, for a better visualization we will take $3$ points, $x$-intercept, $y$-intercept and a point corresponding to a random $x$ value. $2x-y=6$ We have $x$-intercept when $y=0$ $2x-0=6$ $2x=6$ $x=3$ We have first point, $x$-intercept, $B(3,0)$ For $y$-intercept $x=0$ $2\times0-y=6$ $-y=6$ $y=-6$ $y$-intercept is $A(0,-6)$ Now let's write a function of $y$ in terms of $x$, then take any real number and calculate $3^{rd}$ point (Take different value than the intercepts we found earlier). $2x-y=6$ $y=2x-6$ Let $x$ be $2$ $y=2\times2-6$ $y=-2$ We found $C(2,-2)$ To sketch the graph, we simply plot these points and connect them. Also see the image above. The graph is symmetric to only a line perpendicular to this line. $(b)$ $y=-(x+1)^2$ We have $x$-intercept when $y=0$ $0=-(x+1)^2$ $x+1=0$ $x=-1$ We have first point, $x$-intercept, $B(-1,0)$ For $y$-intercept $x=0$ $y=-(0+1)^2$ $y=-1$ $y$-intercept is $A(0,-1)$ In this case we have quadratic function which has a form of parabola. The $-$ sign in front of the equation informs us that the parabola is concave down. And as we have only one $x$-intercept, we can assume that the point is vertex of parabola. Let's find one more point for easier sketching $x=-3$ $y=-(-3+1)^2$ $y=-4$ $C(-3,-4)$ We can now plot the points and approximately connect them, giving a form of parabola. See the image below.
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