The function f is given by:
![\begin{gathered} f(x)=2x^2-x+6 \\ \text{ Rewrite the quadratic function in vertex form} \\ f(x)=2(x^2-(1)/(2)x)+6 \\ =2((x-(1)/(4))^2-(-(1)/(4))^2)+6 \\ =2(x-(1)/(4))^2-2((1)/(16))+6 \\ =2(x-(1)/(4))^2+(47)/(8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vbuiz4zvkwvzyaihzvdg15d7mdobxcqbod.png)
If a quadratic function is written in the form:
![\begin{gathered} a(x-h)^2+k \\ where: \\ a>0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/g52d5gsw93n6i5iu5d3r7p8zncwtvtk8kg.png)
Then the function has a minimum point at (h,k)
And the minimum is k
In this case,
![\begin{gathered} a=2\gt0 \\ h=(1)/(4)=0.25 \\ k=(47)/(8)=5.875 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9mwvs6sy9a931bbo9mxkp307hm03cfz7j6.png)
Therefore, the minimum of the function f is at (0.25, 5.875)
The minimum of the function given by the table is at (-1, -6).
Therefore, the required horizontal distance is given by:
![0.25-(-1)=1.25](https://img.qammunity.org/2023/formulas/mathematics/college/8osp8e14v6330k4gx8ykdued7jxgxlmrbs.png)
Therefore, the horizontal distance is 1.25