The function f is given by:
[tex]\begin{gathered} f(x)=2x^2-x+6 \\ \text{ Rewrite the quadratic function in vertex form} \\ f(x)=2(x^2-\frac{1}{2}x)+6 \\ =2((x-\frac{1}{4})^2-(-\frac{1}{4})^2)+6 \\ =2(x-\frac{1}{4})^2-2(\frac{1}{16})+6 \\ =2(x-\frac{1}{4})^2+\frac{47}{8} \end{gathered}[/tex]
If a quadratic function is written in the form:
[tex]\begin{gathered} a(x-h)^2+k \\ where: \\ a>0 \end{gathered}[/tex]
Then the function has a minimum point at (h,k)
And the minimum is k
In this case,
[tex]\begin{gathered} a=2\gt0 \\ h=\frac{1}{4}=0.25 \\ k=\frac{47}{8}=5.875 \end{gathered}[/tex]
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:
[tex]0.25-(-1)=1.25[/tex]
Therefore, the horizontal distance is 1.25
