2x^2 -12x +23=2*(x^2-6x+23/2)
(x-a)^2= x^2-2ax+a^2
(x-3)^2= x^2-2*3*x+3^2= x^2-6x+3^2
2*(x^2-6x+23/2)=
=2*(x^2-2*3*x +3^2+ 23/2 -3^2)
=2(x^2-6x+3^2)+2*(23/2)-2*(3^2)
=2*(x-3)^2+23-2*9
=2*(x-3)^2+23-18
=2*(x-3)^2 +5
Proof:
2*(x-3)^2+5=2*(x^2-2*3*x+3^2)+5
=2*(x^2-6x+9)+5
=2x^2-12x+18+5=2x^2-12x+23
So:
2x^2 -12x +23 = 2*(x-3)^2 +5
a vertical stretch =2
a y-intercept is a point where the graph of a function intersects with the y-axis
2*0^2-12*0+23=0-0+23=23
vertical translation = 23
Quadratic function have extreme point
for x=-(b/2a)
In this case a=2 b=-12
-(b/2a)=-[(-12/(2*2]=-(-12/4)=12/4=3
2x^2 -12x +23 = 2*(x-3)^2 +5
k=2
p=3
q=23
HELP Not great in these
Express y=2x^2 -12x +23 in the form
y=2(x-c)^2 + d
The graph of y=x^2 is transformed into the graph of y=2x^2 - 12x +23 by the transformation
a vertical stretch with scale factor k followed by,
A horizontal translation of p units followed by,
a vertical translation of q units
what are the values of k, p , and q
HELP PLEASE!
1 answer