Intermediate Algebra (12th Edition)

Published by Pearson

Chapter 8 - Section 8.5 - Graphs of Quadratic Functions - 8.5 Exercises: 9

Answer

$(1,0)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the vertex of the given quadratic function, $f(x)=(x-1)^2 ,$ convert the function in the form $f(x)=a(x-h)^2+k.$ Once in this form, the vertex is located at $(h,k).$ $\bf{\text{Solution Details:}}$ In the form $f(x)=a(x-h)^2+k,$ the function above is equivalent to \begin{array}{l}\require{cancel} f(x)=(x-1)^2+0 .\end{array} In the function above, $h= 1$ and $k= 0 .$ Hence, the vertex is \begin{array}{l}\require{cancel} (1,0) .\end{array}

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