Respuesta :
The equation is
[tex]y=x^2+4x-5[/tex]We can already find the vertex using the vertex formulas
[tex]\begin{gathered} x_V=-\frac{b}{2a}=\frac{-4}{2}=-2 \\ \\ \\ y_V=-\frac{\Delta}{4a}=-\frac{b^2-4ac}{4a}=-\frac{16+20}{4}=-\frac{36}{4}=-9 \end{gathered}[/tex]Therefore the vertex is
[tex](x_V,y_V)=(-2,-9)[/tex]Now we have the vertex we also have the axis of symmetry and the max/min of the function, in that case, it's a minimum because a > 0. Therefore
[tex]\begin{gathered} \text{ axis of symmetry = }x_V=-2 \\ \\ \min\lbrace y\rbrace=y_V=-9 \end{gathered}[/tex]We can find the x-intercept easily
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-4\pm\sqrt{4^2+4\cdot5}}{2} \\ \\ x=\frac{-4\pm\sqrt{16+20}}{2} \\ \\ x=\frac{-4\pm\sqrt{36}}{2} \\ \\ x=\frac{-4\pm6}{2} \\ \\ \end{gathered}[/tex]Hence
[tex]\begin{gathered} x=\frac{-4\pm6}{2}=-2\pm3 \\ \\ x_1=-1 \\ x_2=-5 \end{gathered}[/tex]The y-intercept is just the c value, then it's -5.
Now we can do the domain, there's no restriction for parabolas in the domain, then
[tex]\text{ domain = }\mathbb{R}[/tex]And the range is
[tex]\text{ range = \lbrack}y_V,+\infty)=[-9,+\infty)[/tex]