What is the vertex form of y=11x^2 - 4x + 31 ?

1 Answer
Oct 22, 2017

The vertex form of equation is $y = 11 {\left(x - \frac{2}{11}\right)}^{2} + 30 \frac{7}{11}$
of which vertex is at (2/11 , 30 7/11)

Explanation:

$y = 11 {x}^{2} - 4 x + 31 \mathmr{and} y = 11 \left({x}^{2} - \frac{4}{11} x\right) + 31$ or

$y = 11 \left({x}^{2} - \frac{4}{11} x + {\left(\frac{2}{11}\right)}^{2}\right) - 11 \cdot \frac{4}{11} ^ 2 + 31$ or

$y = 11 {\left(x - \frac{2}{11}\right)}^{2} - \frac{4}{11} + 31$ or

$y = 11 {\left(x - \frac{2}{11}\right)}^{2} + \frac{337}{11}$ or

$y = 11 {\left(x - \frac{2}{11}\right)}^{2} + 30 \frac{7}{11}$

The vertex form of equation is $y = 11 {\left(x - \frac{2}{11}\right)}^{2} + 30 \frac{7}{11}$

of which vertex is at (2/11 , 30 7/11) [Ans]