Answer
See explanation
Work Step by Step
Using a graphing utility, the graph is as shown.
The vertex is $(-2,-3)$.
The axis of symmetry is $x=-2$.
The $x$-intercepts are $(-4.45,0)$ and $(0.45,0)$.
Finding the standard form of the quadratic function:
$$f(x)=\frac{1}{2}(x^2+4x-2)$$ $$f(x)=\frac{1}{2}(x^2+4x)-1$$ $$f(x)=\frac{1}{2}\left(x^2+4x+\left(\frac{4}{2}\right)^2\right)-1-\frac{1}{2}\left(\frac{4}{2}\right)^2$$ $$f(x)=\frac{1}{2}(x^2+4x+4)-1-2$$ $$f(x)=\frac{1}{2}(x+2)^2-3$$
Thus, from the form $y=a(x-h)^2+k$, the vertex is $(-2,-3)$ and axis of symmetry is $x=-2$.
Finding the $x$-intercept by setting $y=0$:
$$0=\frac{1}{2}(x+2)^2-3$$ $$3=\frac{1}{2}(x+2)^2$$ $$\frac{1}{2}(x+2)^2=3$$ $$(x+2)^2=6$$ $$x+2=\pm\sqrt6$$ $$x=-2\pm\sqrt6$$ $$x_1=-2-\sqrt6\approx-4.45$$ $$x_2=-2+\sqrt6\approx0.45$$
Thus, the $x$-intercepts are $(-4.45,0)$ and $(0.45,0)$.
