The equation of the given parabola is
[tex]y=\frac{1}{8}x^2_{}+2x[/tex]
Rewrite the equation in the vertex form
[tex]y=a(x-h)^2+k[/tex]
The equation becomes
[tex]\begin{gathered} y=\frac{1}{8}x^2+2x \\ y=\frac{1}{8}(x^2+16x) \\ 8y=x^2+16x \\ 8y=x^2+6x+64-64 \\ 8y=(x+8)^2-64 \end{gathered}[/tex]
Divide through the equation by 8
This gives
[tex]y=\frac{1}{8}(x+8)^2-8[/tex]
Comparing the equation with the vertex form
It follows
[tex]a=\frac{1}{8},h=-8,k=-8[/tex]
The focus of a parabola in vertex form is given as
[tex]F=(h,k+\frac{1}{4a})[/tex]
Substitute h = -8, k = -8 and a = 1/8 into the formula for focus
This gives
[tex]F=(-8,-8+\frac{1}{4(\frac{1}{8})})[/tex]
Simplify the expression
[tex]\begin{gathered} F=(-8,-8+\frac{1}{\frac{1}{2}}) \\ F=(-8,-8+2) \\ F=(-8,-6) \end{gathered}[/tex]
Therefore, the focus of the parabola is at (-8, -6)
