Answer
Graph:
.
Work Step by Step
$f(x)=x^{2}+2x-3$
$f(x)=ax^{2}+bx+c,\qquad a=1,b=2,c=-3$
Coefficient $a$ determines which way the parabola opens.
Here $a=1$, which is positive, so the parabola opens $\mathrm{u}\mathrm{p}$
Coefficient $c$ tells us the y-intercept.
Here, $(0,-3)$ lies on the graph.
The vertex lies on the axis of symmetry, $x=-\displaystyle \frac{b}{2a}$
$x=-\displaystyle \frac{2}{1}=-1$
At $x=-1,\ f(-1)=(-1)^{2}+2(-1)-3=-4$
$\mathrm{V}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{e}\mathrm{x}$:$\quad (-1,-4)$
For the x-intercepts:
$x^{2}+2x-3=0$
$(x+3)(x-1)=0$
x-intercepts:$\qquad(-3,0),(1,0)$
With this information, we sketch the graph.
