Answer
See explanation
Work Step by Step
Finding the quadratic function in standard form:
$$f(x)=4x^2-4x+21$$ $$f(x)=4(x^2-x)+21$$ $$f(x)=4\left(x^2-x+\left(\frac{-1}{2}\right)^2\right)+21-4\left(\frac{-1}{2}\right)^2$$ $$f(x)=4\left(x^2-x+\frac{1}{4}\right)+21-1$$ $$f(x)=4\left(x-\frac{1}{2}\right)^2+20$$
The sketch of the graph of the function is as shown.
The vertex is: $$\left(\frac{1}{2},20\right)$$
The axis of symmetry is: $$x=\frac{1}{2}$$
There is no $x$-intercept.
