Question

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

Answer 1

Answer:

The equation in standard form is:

$f(x)=6(x-\frac{1}{12})^2-\frac{23}{24}$

Step-by-step explanation:

Given function:

$f(x)=6x^2-x+1$

We need to convert this in standard form which is given by:

$f(x)=a(x-h)^2+k$

where $a$ represent co-efficient of leading term which is $x^2$

and $(h,k)$ is the vertex (minimum and maximum point) of the curve.

$h$ can be found out using formula $h=\frac{-b}{2a}$

$a=6\ and\ b=-1$

$h=\frac{-(-1)}{2(6)}=\frac{1}{12}$

We can find $k$ by finding $f(h)$ as $k=f(h)$.

$k=f(\frac{1}{12})=6(\frac{1}{12})^2-\frac{1}{12}+1$

$k=(6\times \frac{1}{144})-\frac{1}{12}+1$

$k=\frac{1}{24}-\frac{1}{12}+1$

$k=\frac{1-2+24}{24}=\frac{23}{24}$

Thus the equation in standard form is:

$f(x)=6(x-\frac{1}{12})^2-\frac{23}{24}$