Answered: Hi so my hw has a question where I had…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Hi so my hw has a question where I had to rewrite 2x² -4x-3 in the form of f(x)=a(x-h)² +k which I got someone on here to solve to double check that I was correct and we both got f(x)= 2(x-1)² -5. Making the vertex (1,-5).

^{Now the question I have is in regards to the fourth part of this question which is to chek that solution using:}

^{a. -b/2a=}

^{b. f(-b/2)=}

^{If anyone could tell me how to set each of these up I would be greatful.}

^{Thank you!}

Expert Solution

Given,

$f (x) = 2 x^{2} - 4 x - 3$

On simplification, we get

$\begin{array}{rcl} 2 x^{2} - 4 x - 3 & = & 2 (x^{2} - 2 x) - 3 \\ = & 2 (x^{2} - 2 x + 1 - 1) - 3 \\ = & 2 (x^{2} - 2 x + 1) - 2 - 3 \\ = & 2 {(x - 1)}^{2} - 5 \end{array}$

Now comparing with $f (x) = a {(x - h)}^{2} + k$ , we get

$h = 1 & k = - 5$ .

Therefore vertex is,

$(h, k) = (1, - 5)$

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