Question 478335: Can you please help with the following problem?
Find the vertex f(x) = -2x^2+2x+6
Can you please show the steps so that I can figure out where I am going wrong. Thank you
Found 3 solutions by lwsshak3, Alan3354, stanbon:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertex f(x) = -2x^2+2x+6
**
Standard form of an equation for a parabola: A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. A is a multiplier which affects the steepness of the curve.
..
f(x)=-2x^2+2x+6
completing the square to see equation in standard form
f(x)=-2(x^2-x+1/4)+6+1/2
f(x)=-2(x-1/2)^2+13/2
This is an equation of a parabola with vertex at (1,13/2). The negative coefficient of the lead term means it opens downward, that is, it has a maximum.
see the graph below as a visual check on the answer
..
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Find the vertex f(x) = -2x^2+2x+6
------------
The Line of Symmetry is x = -b/2a
x = -2/(-4) = 1/2
The vertex is (1/2,f(1/2))
f(1/2) = -2*(1/4) + 2*(1/2) + 6 = 13/2
Vertex at (1/2,13/2) or (0.5,6.5)
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! ): Can you please help with the following problem?
Find the vertex f(x) = -2x^2+2x+6
Can you please show the steps so that I can figure out where I am going wrong
------------
Complete the square:
-2x^2+2x+6 = y
-2(x^2-x+?) = y-6-2*?
-2(x^2-x+(1/4)) = y-6-2(1/4)
-2(x-(1/2))^2 = y-(13/2)
-----
Vertex: (1/2,13/2)
----------------------------
Use x = -b/2a
-b/2a = -2/(2(-2) = 1/2
---
f(1/2) = -2(1/2)^2+2(1/2)+ 6
f(1/2) = -1/2+1+6
f(1/2) = 13/2
--------
Vertex: (1/2,13/2)
=======================
Cheers,
Stan H.
===========
|
|
|
