College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 3, Polynomial and Rational Functions - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 288: 24

Answer

a) In standard form, $f(x) = 3(x+\frac{1}{3})^2-\frac{7}{3}$. b) Vertex = $(-\frac{1}{3} -\frac{7}{3})$; x-intercepts are $(\frac{-1-\sqrt{7}}{3}, 0) , (\frac{-1+\sqrt{7}}{3}, 0)$; y-intercept is $(0,-2 )$. c) Graph is attached. d) The domain is all real numbers; the range is $y\ge-\frac{7}{3}$.
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Work Step by Step

a) To find the standard form of some function $f(x) = ax^2+bx+c$, the standard form would be $f(x) = a(x-h)^2+k$ where $h = -\frac{b}{2a}$ and $k = f(h)$. This is a standard result derived in the book simplifies the algebra and gives a closed form for the end result. For this problem, $a= 3,b = 2, c = -2$. Plugging above, we get $h = -\frac{1}{3}, k = f(-\frac{1}{3}) = -\frac{7}{3}$ and hence $f(x) = 3(x+\frac{1}{3})^2-\frac{7}{3}.$ b) The vertex can be easily deduced from the standard form; it is the point $(h, k)$ so in this case $(-\frac{1}{3}, -\frac{7}{3})$. For the x-intercepts, by definition, we know that $y = 0$ hence we say $f(x) = 0 = a(x-h)^2+k = 0$. Solving the above, we get $(x-h)^2 = -\frac{k}{a}\to |x-h|=\sqrt{-\frac{k}{a}} \to x=h\pm\sqrt{-\frac{k}{a}}$.We note that the result is undefined if the ratio $\frac{k}{a}$ is positive. Plugging the above values, we get the two intercepts to be $(\frac{-1-\sqrt{7}}{3}, 0) , (\frac{-1+\sqrt{7}}{3}, 0)$. For the y-intercept, we plug $x=0$ and simplify in the standard form to get $f(0) = a(0-h)^2+k = ah^2+k.$ Plugging in $h$ and $k$, we get the y-intercept to be $(0, -2)$. c)To plot the graph, we plot the above points and join them using a smooth curve, we either get a parabola pointing upwards or downwards (the direction/orientation would be explained in d). d) Polynomial functions, and by extension, quadratic functions, are defined for all real numbers and so the domain is all real numbers. For the range, we would look at the sign of the co-efficient of the largest term (in this case $a$): - if $a>0$, this tells us the function will grow towards positive infinity and hence, the function has a minimum so its range is of the form $y\ge k$ - if $a<0$, this tells us that the function will grows towards negative infinity and hence, the function has a maximum so its range is of the form $y\le k$. Looking at $a$ in this case and plugging in $k$ tells us that the range of this function is $y\ge-\frac{7}{3}$.
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