Answer
See explanation
Work Step by Step
Finding the quadratic function in standard form:
$$f(x)=-x^2-4x+1$$ $$f(x)=-(x^2+4x)+1$$ $$f(x)=-\left(x^2+4x+\left(\frac{4}{2}\right)^2\right)+1+\left(\frac{4}{2}\right)^2$$ $$f(x)=-(x^2+4x+4)+1+4$$ $$f(x)=-(x+2)^2+5$$
The sketch of the graph of the function is as shown.
The vertex is: $$\left(-2,5\right)$$
The axis of symmetry is: $$x=-2$$
The $x$-intercepts are: $$(-4.24,0),(0.24,0)$$
