# What is the vertex form of y=(-x+12)(2x-5)?

Aug 10, 2017

Equation in vertex form is $- 2 {\left(x - \frac{29}{4}\right)}^{2} + \frac{361}{8}$ and vertex is $\left(\frac{29}{4} , \frac{361}{8}\right)$ or $\left(7 \frac{1}{4} , 45 \frac{1}{8}\right)$.

#### Explanation:

This is the intercept form of equation of a parabola as the two intercept on $x$-axis are $12$ and $\frac{5}{2}$. To convert it in vertex form we should multiply RHS and convert it to form $y = a {\left(x - h\right)}^{2} + k$ and vertex is $\left(h , k\right)$. This can be done as follows.

$y = \left(- x + 12\right) \left(2 x - 5\right)$

= $- 2 {x}^{2} + 5 x + 24 x - 60$

= $- 2 \left({x}^{2} - \frac{29}{2} x\right) - 60$

= -2(x^2-2×29/4×x+(29/4)^2)+(29/4)^2×2-60

= $- 2 {\left(x - \frac{29}{4}\right)}^{2} + \frac{841}{8} - 60$

= $- 2 {\left(x - \frac{29}{4}\right)}^{2} + \frac{361}{8}$

and hence vertex is $\left(\frac{29}{4} , \frac{361}{8}\right)$ or $\left(- 7 \frac{1}{4} , 45 \frac{1}{8}\right)$.

graph{y-(-x+12)(2x-5)=0 [0, 20, 0, 50]}