Answer
See image
Work Step by Step
$ \begin{array}{llll}
\text{Function:} & f(x)=x^{2}+x-1 & & \\
\text{coefficients} & a =+1, b=1, c=-1 & & \\
\text{coefficient a} & \text{positive, opens up} & \\ & \\
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a)) & \\
& x_{V}=-1/2=-0.5 & y_{V}=-5/4=-1.25 & \\
& \bf V(-0.5,-1.25) & & \\\\
\text{Line of symmetry} & x=-0.5\\ & & \\
\text{y intercept} & c=-1, & \bf(0,-1) & \\
\text{point symmetric to } & & & \\
\text{the y-intercept} & x=2x_{V}=1 & \bf(1,1)\\ & \\
\text{Zeros:} & f(x)=0 & & \\
& x^{2}+x-1=0 & \text{... quadratic formula} & \\
& \mathrm{x}=\frac{-\mathrm{b}\pm\sqrt{\mathrm{b}^{2}-4\mathrm{a}\mathrm{c}}}{2\mathrm{a}} & & \\
& x=-\frac{1}{2}\pm\frac{\sqrt{5}}{2} & & \\
& \bf\approx(-1.618,0), (0.618,0) & &
\end{array}$
Additional points: $(1,1), (-2,1).$
Using the above information, plot the points and join with a smooth curve (parabola).