convert 3X2-9X+4 to this form a(x+k)2+m

Question

I know the answer to this is 3(x-3/2)2-11/4  The numbers two in the problem are squared. please show how to convert and to expression in the form. thank you

Answer 1

a(x+k)^2+m

=a(x^2+2kx+k^2)+m

=ax^2+2akx+ak^2+m

 

Compare to 3x^2-9x+4

 

3=a

-9=2ak

4=ak^2+m

 

a=3

k=-9/(2a)=-3/2

m=4-ak^2=4-3(-3/2)^2=16/4 - 27/4 = -11/4

 

So 3x^2-9x+4 = 3(x-3/2)^2-11/4

 

GeoGebra check:

http://www.wyzant.com/resources/files/262096/finding_vertex_form

 

Answer 2

Hi Roseland;

3X²-9X+4

This equation is in the format of...

ax²+bx+c

The x-value of the vertex of such is -b/2a.

- {-9/[(2)(3)]}

9/6

x=3/2

y=3X²-9X+4

y=[(3)(3/2)²]-[(9)(3/2)]+4

y=[(3)(9/4)]-(27/2)+4

y=(27/4)-(27/2)+4

y=(27/4)-(54/4)+(16/4)

y=-11/4

vertex is (3/2, -11/4)

Parabolic formula which highlights vertex is...

y=a(x-k)²+m

Vertex is (k,m)

y=3[x-(3/2)]²-11/4

Let's check our work.  Let's FOIL [x-(3/2)]²...(x-3/2)(x-3/2)...

FIRST...(x)(x)=x²

OUTER...(x)(-3/2)=(-3/2)x

INNER...(-3/2)(x)=(-3/2)x

LAST...(-3/2)(-3/2)=9/4

x²-(3/2)x-(3/2)x+(9/4)

x²-3x+(9/4)

y=3[x²-3x+(9/4)]-11/4

y=3x²-9x+(27/4)-(11/4)

y=3x²-9x+(16/4)

y=3x²-9x+4