What is the vertex form of 2y = 7x^2-5x+7?

Nov 2, 2017

Vertex form of equation is $y = \frac{7}{2} {\left(x - \frac{5}{14}\right)}^{2} + 3 \frac{3}{56}$

Explanation:

$2 y = 7 {x}^{2} - 5 x + 7 \mathmr{and} y = \frac{7}{2} {x}^{2} - \frac{5}{2} x + \frac{7}{2}$ or

$y = \frac{7}{2} \left({x}^{2} - \frac{5}{7} x\right) + \frac{7}{2}$ or

$y = \frac{7}{2} \left\{{x}^{2} - \frac{5}{7} x + {\left(\frac{5}{14}\right)}^{2}\right\} - \frac{7}{2} \cdot {\left(\frac{5}{14}\right)}^{2} + \frac{7}{2}$ or

$y = \frac{7}{2} \left\{{x}^{2} - \frac{5}{7} x + {\left(\frac{5}{14}\right)}^{2}\right\} - \frac{25}{56} + \frac{7}{2}$

$y = \frac{7}{2} {\left(x - \frac{5}{14}\right)}^{2} + \frac{171}{56}$. Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = \frac{5}{14} , k = \frac{171}{56} \mathmr{and} k = 3 \frac{3}{56}$

So vertex is at $\left(\frac{5}{14} , 3 \frac{3}{56}\right)$ and vertex form of

equation is $y = \frac{7}{2} {\left(x - \frac{5}{14}\right)}^{2} + 3 \frac{3}{56}$ [Ans]