Answer: The required vertex of the given function is \(\left(\dfrac{1}{2},\dfrac{7}{4}\right).\)
Step-by-step explanation: Given that Marsha wants to determine the vertex of the following quadratic function :
\(f(x)=x^2-x+2~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\)
We are to find the vertex of the function (i).
We know that
the form of a vertex function g(x) is as follows :
\(g(x)=a(x-h)^2+k,\) where (h, k) is the vertex.
From equation (i), we have
\(f(x)\\\\=x^2-x+2\\\\=\left(x^2-2\imes x\times\dfrac{1}{2}\right)+2\\\\\\=\left(x^2-2\imes x\times\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+2-\left(\dfrac{1}{2}\right)^2\\\\\\=\left(x-\dfrac{1}{2}\right)^2+2-\dfrac{1}{4}\\\\\\=\left(x-\dfrac{1}{2}\right)^2+\dfrac{8-1}{4}\\\\\\=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}.\)
Comparing the above with the vertex form of a function, we get
the vertex of the function f(x) is \(\left(\dfrac{1}{2},\dfrac{7}{4}\right).\)
Thus, the required vertex of the given function is \(\left(\dfrac{1}{2},\dfrac{7}{4}\right).\)
