What is the vertex form of #y=(2x-9)(3x-1)#?

Question

Isaiah Anthony · Accepted Answer

#y=6(x-29/24)^2-625/24#

Given - #y=(2x-9)(3x-1)# Vertex form of the equation is - #y=a(x-h)^2+k# Find the vertex first - #y=6x^2-27x-2x+9# #y=6x^2-29x+9# x coordinate of the vertex #x=(-b)/(2a)# #x=(-(-29))/(2xx6)=29/12# y-coordinate of the vertex #y=6(29/12)^2-29(29/12)+9# #y=6(841/144)-841/12+9# #y=5046/144-841/12+9# #y=(5046-10092+1296)/144=-3750/144=-625/24# Vertex #(29/12, -625/24)# #a=6# [coefficient of #x^2#] #h=29/12# [x coordinae of the vertex] #k=-625/24# The vertex form of the parabola equation is - #y=6(x-29/24)^2+((-625)/24)# #y=6(x-29/24)^2-625/24#

Isaiah Anthony · Answer

undefined To find the vertex form of $ y = (2x - 9)(3x - 1) $, you first need to expand the equation:

$$ y = (2x - 9)(3x - 1) $$
$$ y = 6x^2 - 2x - 27x + 9 $$
$$ y = 6x^2 - 29x + 9 $$

Next, complete the square to write the quadratic equation in vertex form $ y = a(x - h)^2 + k $:

Starting with $ y = 6x^2 - 29x + 9 $:

Factor out the leading coefficient from the first two terms:
$$ y = 6(x^2 - \frac{29}{6}x) + 9 $$

To complete the square inside the parentheses, take half of the coefficient of $ x $ and square it:
$$ \left(\frac{-29}{6 \times 2}\right)^2 = \left(\frac{-29}{12}\right)^2 = \frac{841}{144} $$

Add and subtract this value inside the parentheses:
$$ y = 6\left(x^2 - \frac{29}{6}x + \frac{841}{144} - \frac{841}{144}\right) + 9 $$

Simplify inside the parentheses:
$$ y = 6\left(\left(x - \frac{29}{12}\right)^2 - \frac{841}{144}\right) + 9 $$

Combine like terms:
$$ y = 6\left(x - \frac{29}{12}\right)^2 - \frac{841}{24} + 9 $$

Convert $-\frac{841}{24}$ to a common denominator of 24:
$$ y = 6\left(x - \frac{29}{12}\right)^2 - \frac{841}{24} + \frac{216}{24} $$
$$ y = 6\left(x - \frac{29}{12}\right)^2 - \frac{625}{24} $$

So, the vertex form of $ y = (2x - 9)(3x - 1) $ is:
$$ y = 6\left(x - \frac{29}{12}\right)^2 - \frac{625}{24} $$