Question #b0cbf

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Question #b0cbf

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Answer 1 · 2017-05-21T00:10:38

$f (x) = {(x - 1)}^{2} + 4$ $f(x)=(x-1)^2+4$
Minimum value is 4.

Explanation:

$f (x) = x^{2} - 2 x + 1 + 4$ $f(x)=x^2-2x+1+4$
$f (x) = {(x - 1)}^{2} + 4$ $f(x)=(x-1)^2+4$

The graph of this function is very similar to the graph of $f (x) = x^{2}$ $f(x) = x^2$ . $f (x) = {(x - 1)}^{2} + 4$ $f(x)=(x-1)^2+4$ would just be $f (x) = x^{2}$ $f(x)=x^2$ shifted one unit to the right and four units up.

The maximum/minimum point can be found using the vertex. Since the graph of this function faces up, you need to solve for the minimum point.

The vertex could be found by:
$x = - \frac{b}{2 a}$ $x=-b/(2a)$ where the function is expressed as $f (x) = a x^{2} + b x + c$ $f(x)=ax^2+bx+c$ .
$x = - \frac{- 2}{2 \cdot 1}$ $x=-(-2)/(2*1)$
$x = 1$ $x=1$
$f (1) = 4$ $f(1)=4$
The minimum value is 4.

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