Answer
See explanation
Work Step by Step
Finding the quadratic function in standard form:
$$f(x)=2x^2-x+1$$ $$f(x)=2\left(x^2-\frac{1}{2}x\right)+1$$ $$f(x)=2\left(x^2-\frac{1}{2}x+\left(\frac{-\frac{1}{2}}{2}\right)^2\right)+1-2\left(\frac{-\frac{1}{2}}{2}\right)^2$$ $$f(x)=2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+1-\frac{1}{8}$$ $$f(x)=2\left(x-\frac{1}{4}\right)^2+\frac{7}{8}$$
The sketch of the graph of the function is as shown.
The vertex is: $$\left(\frac{1}{4},\frac{7}{8}\right)$$
The axis of symmetry is:
$$x=\frac{1}{4}$$
There is no $x$-intercept.
