SOLUTION: Given that f(x) = 2x^2 - 8x + 1. Express 2x^2 - 8x + 1 in the form a(x+b)^2 + c,where a and b are integers. Find the coordinates of the stationary point on the graph of y = f(x).

Algebra ->  Test -> SOLUTION: Given that f(x) = 2x^2 - 8x + 1. Express 2x^2 - 8x + 1 in the form a(x+b)^2 + c,where a and b are integers. Find the coordinates of the stationary point on the graph of y = f(x).      Log On


   

Question 1184797: Given that f(x) = 2x^2 - 8x + 1. Express 2x^2 - 8x + 1 in the form a(x+b)^2 + c,where a and b are integers. Find the coordinates of the stationary point on the graph of y = f(x).
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29+=+2x%5E2+-+8x+%2B+1.........complete square
f%28x%29+=+%282x%5E2+-+8x%29+%2B+1
f%28x%29+=+2%28x%5E2+-+4x%2Bb%5E2%29-2b%5E2+%2B+1..........b=4%2F2=2
f%28x%29+=+2%28x%5E2+-+4x%2B2%5E2%29-2%2A2%5E2+%2B+1
f%28x%29+=+2%28x+-+2%29%5E2-8+%2B+1
f%28x%29+=+2%28x+-+2%29%5E2-7
vertex is at (2,-7)
the stationary point on the graph:
The graph of a quadratic function (ie a parabola) only has a single stationary point.
For an ‘up’ parabola this is the minimum; for a ‘down’ parabola it is the maximum (no need to talk about ‘local’ here) The y+value of the stationary point is thus the minimum or maximum+value of the quadratic function
so, the stationary point on the graph is minimum (2,-7)
or,
The specific nature of a stationary point at x can in some cases be determined by examining the second derivative f''%28x%29:
f'%28x%29=2x-8
f''%28x%29=2
y=2%2A2%5E2+-+8%2A2+%2B+1
y=+-+7
the stationary point on the graph is (2,-7)