SOLUTION: Find the coordinates of the turning point of the curve {{{y = 6 + 4x - x^2}}} (by expressing this equation in the form {{{a - (x + b)^2}}} ) and determine the nature of this turnin

Algebra ->  Coordinate-system -> SOLUTION: Find the coordinates of the turning point of the curve {{{y = 6 + 4x - x^2}}} (by expressing this equation in the form {{{a - (x + b)^2}}} ) and determine the nature of this turnin      Log On


   

Question 474084: Find the coordinates of the turning point of the curve y+=+6+%2B+4x+-+x%5E2 (by expressing this equation in the form a+-+%28x+%2B+b%29%5E2 ) and determine the nature of this turning point.
*Please answer as soon as possible bro :)
Found 2 solutions by nerdybill, ewatrrr:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+6+%2B+4x+-+x%5E2
"turning point" is at the vertex, where the x coordinate is at:
x = -b/(2a)
x = -4/(2(-1))
x = -4/(-2)
x = 2
.
find y by plugging it into equation:
y+=+6+%2B+4x+-+x%5E2
y+=+6+%2B+4%282%29+-+2%5E2
y+=+6+%2B+8+-+4
y+=+10
.
turning point is at (2,10)
Since the coefficient associated with the x^2 is negative, it is a parabola that opens downwards.
This means (2,10) is the peak.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Find the coordinates of the turning point of the curve 

y+=+-x%5E2+%2B4x%2B6  |Turning point is the Vertex of the Parabola

  y = -[x^2 - 4x] + 6

  y = -[(x-2)^2 - 4] + 6

  y = -(x-2)^2 + 10  V(2,10), the turning point

       a = -1 -1<0, Parabola opens downward