You can
put this solution on YOUR website!I need to find the axes, roots, and the vertices of the 3 following questions. Can you please show it to me in the completing the square method.
The questions are
y=2x^2-x+6
2x^2 - x + ? = y-6+?
2(x^2 - (1/2)x + (1/4)^2 = y - 6 + 2(1/4)^2
2(x- (1/4))^2 = y - (48/8) + (1/8)
2(x-(1/4)^2 = y - (47/8)
(x - (1/4)^2 = (1/2)(y -(47/8))
------------------------------------
Vertex: (1/4 , (-47/8))
vertical axis: x = 1/4
Roots: No Real roots
Use Quadratic formula to find the complex roots.
-----------------------------------
y=x^2-8x+2
x^2 - 8x = y-2
x^2 - 8x + 16 = y -2 + 16
(x-4)^2 = y + 14
----------------------
Vertex: (4 , -14)
vertecal axis: x = 4
Roots: Use the quadratic formula.
---------------------------
y=4x^2-x+1
4x^2 - x = y - 1
4(x^2 - (1/4)x + (1/8)^2) = y - 1 + 4*(1/8)^2
4(x - (1/8))^2 = y - 15/16
(x-(1/8))^2 = (1/4)(y - 15/16)
----
Vertex: (1/8 , 15/16)
vertical axis: x = 1/8
Roots: Use the Quadratic formula
=====================================
Cheers,
Stan H.
You can
put this solution on YOUR website!The x-co-ordinate,
, of the vertex is exactly midway
between the roots, so if the roots are
and
the vertex is at (
,
)
Another way to find the vertex is, if the equation is in
the form
, then
is at
. I'll solve both ways:
(1)
To complete the square, take 1/2 the co-efficient of
,
square it, and add it to both sides.
First set equation equal to
and subtract
from both sides
Divide both sides by
Take the square root of both sides
The 2 roots are:
and
Now I find the point midway between the roots
(notice the terms with
cancel)
And now the easy way:
Now just plug this into the equation to find
I'll plot it
(