Write the quadratic function in standard form.f(x) = 6x^2 − x + 1

Answer:The equation in standard form is:[tex]f(x)=6(x-\frac{1}{12})^2-\frac{23}{24}[/tex]Step-by-step explanation:Given function:[tex]f(x)=6x^2-x+1[/tex]We need to convert this in standard form which is given by:[tex]f(x)=a(x-h)^2+k[/tex]where [tex]a[/tex] represent co-efficient of leading term which is [tex]x^2[/tex]and [tex](h,k)[/tex] is the vertex (minimum and maximum point) of the curve.[tex]h[/tex] can be found out using formula [tex]h=\frac{-b}{2a}[/tex][tex]a=6\ and\  b=-1[/tex][tex]h=\frac{-(-1)}{2(6)}=\frac{1}{12}[/tex]We can find [tex]k[/tex] by finding [tex]f(h)[/tex] as [tex]k=f(h)[/tex].[tex]k=f(\frac{1}{12})=6(\frac{1}{12})^2-\frac{1}{12}+1[/tex][tex]k=(6\times \frac{1}{144})-\frac{1}{12}+1[/tex][tex]k=\frac{1}{24}-\frac{1}{12}+1[/tex][tex]k=\frac{1-2+24}{24}=\frac{23}{24}[/tex]Thus the equation in standard form is:[tex]f(x)=6(x-\frac{1}{12})^2-\frac{23}{24}[/tex]