Answer
See explanation
Work Step by Step
Using a graphing utility, the graph is as shown.
The vertex is $(-3,-8.4)$.
The axis of symmetry is $x=-3$.
The $x$-intercepts are $(-6.74,0)$ and $(0.74,0)$.
Finding the standard form of the quadratic function:
$$f(x)=\frac{3}{5}(x^2+6x-5)$$ $$f(x)=\frac{3}{5}(x^2+6x)-3$$ $$f(x)=\frac{3}{5}\left(x^2+6x+\left(\frac{6}{2}\right)^2\right)-3-\frac{3}{5}\left(\frac{6}{2}\right)^2$$ $$f(x)=\frac{3}{5}(x^2+6x+9)-3-\frac{27}{5}$$ $$f(x)=\frac{3}{5}(x+3)^2-\frac{42}{5}$$ $$f(x)=\frac{3}{5}(x+3)^2-8.4$$ Thus, from the form $y=a(x-h)^2+k$, the vertex is $(-3,-8.4)$ and axis of symmetry is $x=-3$.
Finding the $x$-intercept by setting $y=0$:
$$0=\frac{3}{5}(x+3)^2-8.4$$ $$8.4=\frac{3}{5}(x+3)^2$$ $$\frac{3}{5}(x+3)^2=8.4$$ $$(x+3)^2=14$$ $$x+3=\pm\sqrt{14}$$ $$x=-3\pm\sqrt{14}$$ $$x_1=-3-\sqrt{14}\approx-6,74$$ $$x_2=-3+\sqrt{14}\approx0.74$$ Thus, the $x$-intercepts are $(-6.74,0)$ and $(0.74,0)$.
