# What is the vertex of  y= 3x^2-x+7(x-5)^2 ?

May 6, 2018

Vertex is $\left(3 \frac{11}{20} , 48 \frac{39}{40}\right)$

#### Explanation:

Let us convert this to vertex form $y = a {\left(x - h\right)}^{2} + k$, which has$\left(h , k\right)$ as vertex

$y = 3 {x}^{2} - x + 7 {\left(x - 5\right)}^{2}$

= $3 {x}^{2} - x + 7 \left({x}^{2} - 10 x + 25\right)$

= $3 {x}^{2} - x + 7 {x}^{2} - 70 x + 175$

= $10 {x}^{2} - 71 x + 175$

= $10 \left({x}^{2} - \frac{71}{10} x\right) + 175$

= $10 \left({x}^{2} - 2 \cdot \frac{71}{20} \cdot x + {\left(\frac{71}{20}\right)}^{2} - {\left(\frac{71}{20}\right)}^{2}\right) + 175$

= $10 \left({x}^{2} - 2 \cdot \frac{71}{20} \cdot x + {\left(\frac{71}{20}\right)}^{2}\right) - 10 {\left(\frac{71}{20}\right)}^{2} + 175$

= $10 {\left(x - \frac{71}{20}\right)}^{2} + \frac{1959}{40}$

= $10 {\left(s - 3 \frac{11}{20}\right)}^{2} + 48 \frac{39}{40}$

Hence vertex is $\left(3 \frac{11}{20} , 48 \frac{39}{40}\right)$