What is the vertex form of y=x^2-5x+4 ?

1 Answer
Mar 2, 2016

$y = {\left(x - \frac{5}{2}\right)}^{2} - \frac{9}{4}$

Explanation:

the standard form of a quadratic function is $a {x}^{2} + b x + c$

the function $y = {x}^{2} - 5 x + 4 \text{ is in this form }$

by comparison: a = 1 , b = - 5 and c = 4

the vertex form of the function is $y = {\left(x - h\right)}^{2} + k$

where (h,k) are the coords of the vertex.

x-coord (h) = $\frac{- b}{2 a} = - \frac{- 5}{2} = \frac{5}{2}$

and y-coord ( k ) = ${\left(\frac{5}{2}\right)}^{2} - 5 \left(\frac{5}{2}\right) + 4 = - \frac{9}{4}$

here ( h, k) = ($\frac{5}{2} , - \frac{9}{4} \text{) and } a = 1$

$\Rightarrow y = {\left(x - \frac{5}{2}\right)}^{2} - \frac{9}{4} \text{ is the equation }$