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In Exercises \(19-30\) , find the extreme values of the function and where they occur. \(y=2 x^{2}-8 x+9\)

Short Answer

Expert verified
The extreme value of the function is 1 and it occurs at \(x=2\).

Step by step solution

01

Find the x-coordinate of the Vertex

The x-coordinate of the vertex of a parabola given by \(y=ax^{2}+bx+c\) is \(-b/2a\). Here, \(a = 2\) and \(b = -8\), hence the x-coordinate of the vertex is \((-(-8))/(2*2) = 2\).
02

Find the y-coordinate of the Vertex

Now plug the value \(x=2\) into the original function to find the y-coordinate of the vertex. It will be \(y=2(2)^2-8*2+9 = 1\). Therefore, the y-coordinate of the vertex (minimum value of y) is 1.
03

Interpret the Result

The minimum value of the function is 1 and it occurs at \(x=2\). The function will get larger as x moves away from 2 on either side, implying that \(y=1\) is indeed the minimum value of the function.

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