SOLUTION: Find the coordinates of the vertexfor the parabola defined by the given quadraric funtion. f(x)= -x^2-2x+1

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find the coordinates of the vertexfor the parabola defined by the given quadraric funtion. f(x)= -x^2-2x+1      Log On


   

Question 358372: Find the coordinates of the vertexfor the parabola defined by the given quadraric funtion. f(x)= -x^2-2x+1
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The vertex is half way between the 2 roots
f%28x%29+=+-x%5E2+-+2x+%2B+1
Set f%28x%29+=+0
+-x%5E2+-+2x+%2B+1+=+0
Using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+-1
b+=+-2
c+=+1
x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A%28-1%29%2A1+%29%29%2F%282%2A%28-1%29%29+
x+=+%282+%2B-+sqrt%28+4+%2B+4+%29%29%2F+-2%29+
x+=+%282+%2B-+sqrt%28+8+%29%29%2F+-2%29+
x+=+%282+%2B-+2%2Asqrt%28+2+%29%29%2F+-2%29+
x+=+-1+%2B+sqrt%282%29
and
x+=+-1+-+sqrt%282%29
x+=+-1 is half way between these 2 roots
Actually, this equals -b%2F%282a%29 and
-b%2F%282a%29+=+-%28-2%29%2F%282%2A%28-1%29%29
+-%28-2%29%2F%282%2A%28-1%29%29+=+-1
Now plug x+=+-1 into equation:
f%28x%29+=+-x%5E2+-+2x+%2B+1
f%28-1%29+=+-%28-1%29%5E2+-+2%2A%28-1%29+%2B+1
f%28-1%29+=+-1+%2B+2+%2B+1
f%28-1%29+=+2
The vertex is at (-1,2)
Here's a plot:
+graph%28+400%2C+400%2C+-5%2C+5%2C+-5%2C+5%2C+-x%5E2+-+2x+%2B+1%29+