Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 60


The prove is as shown below.

Work Step by Step

We have parabola $y=ax^2+bx+c$. The square completion is done as follows: Step 1: Take coefficient of $x^2$ common throughout. $y=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)$ Step 2: Add and subtract $\left(\dfrac{\text{Coefficient of x}}{2}\right)^2=\left(\dfrac{b}{2a}\right)^2$ inside the parenthesis. $\Rightarrow y=a\left(x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}\right)$ Step 3: Multiply and divide term with $x$ by 2. $\Rightarrow y=a\left(x^2+2\cdot \dfrac{b}{2a}\cdot x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}\right)$ Step 4: Complete square and simplify. $\Rightarrow y=a\left(\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2+\dfrac{c}{a}\right)$ $\Rightarrow y=a\left(\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2}+\dfrac{c}{a}\right)$ $\Rightarrow y=a\left(\left(x+\dfrac{b}{2a}\right)^2+\dfrac{-b^2+4ac}{4a^2}\right)$ $\Rightarrow y=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{-b^2+4ac}{4a}$ Thus, equation is of the form $y=a(x+h)^2+k$, where $h=\dfrac{b}{2a}$ and $k=\dfrac{-b^2+4ac}{4a}$, which is congruent to $y=ax^2$ with horizontal translation by $|h|$ units and vertical translation by $|k|$ units. Hence, proved.
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