How do you write # y =5x^2 - 3x + 2# in vertex form?

2 Answers
Nov 5, 2017

#y=5(x-3/10)^2+31/20#

Explanation:

Vertex form: #y=a(x-h)^2+k#

#y=5x^2-3x+2#

#y=5(x^2-3/5x)+2#

#y=5(x^2-2*3/10x+9/100)+2-5*9/100#

#y=5(x-3/10)^2+2-9/20#

#y=5(x-3/10)^2+31/20#

Nov 9, 2017

#y= 5(x-3/10)^2 +31/20#

Explanation:

Vertex form is a useful way of writing the equation of a parabola because it tells you the co-ordinates of the turning point, (the vertex).

It is written as #y = p(x+q)^2 +r#
The vertex is at #(-q,r)#

#y = 5x^2-3x+2" "larr# in the form #ax^2+bx+c#

#y = 5[x^2-3/5x" "+2/5]" "larr# you must have #1x^2#

Now you need to add in a term to create the square of the binomial.
The missing term comes from #(b/2)^2# where #b = (-3/5)#

#(-3)/5 div 2 =(-3)/10#

If you add an extra term, you must subtract it as well to keep the equation the same.

#y = 5[x^2-3/5x + color(green)(((-3)/10)^2)+2/5 -color(green)(((-3)/10)^2]#

#y = 5[x^2-3/5x +9/100+2/5 -9/100]#

#y = 5[(color(blue)(x^2-3/5x +9/100))+(color(red)(2/5 -9/100))]#

#y = 5[color(blue)((x-3/10))^2+(color(red)(40/100 -9/100))]#

#y = 5[color(blue)((x-3/10))^2+(color(red)(31/100))]" "larr# distribute the #5#

#y = 5color(blue)((x-3/10))^2+5(color(red)(31/100))#

#y= 5(x-3/10)^2 +31/20#

The vertex is #(3/10,31/20)#