The given equation is
[tex]y=\frac{1}{4}x^2+x-6[/tex]
The vertex has the form V(h,k), where
[tex]h=-\frac{b}{2a},k=f(h)[/tex]
Where a = 1/4 and b = 1.
[tex]\begin{gathered} h=-\frac{1}{2(\frac{1}{4})}=-\frac{1}{\frac{1}{2}}=-2 \\ k=f(-2)=\frac{1}{4}(-2)^2+(-2)-6=\frac{1}{4}\cdot4-2-6 \\ k=1-8=-7 \end{gathered}[/tex]
Hence, the vertex is (-2,-7).
The axis of symmetry is x = -2.
The y-intercept would be -6 because that's the independent term.
The graph of this equation would be
