SOLUTION: For what value of c will the minimum value of the function f(x)= x^2+2x+c be sqrt(2)?

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Question 1064461: For what value of c will the minimum value of the function f(x)= x^2+2x+c be sqrt(2)?
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Lead coefficient is 1, a positive value, so the vertex will be the minimum value for f(x).

f%28x%29=x%5E2%2B2x%2Bc
f%28x%29=x%5E2%2B2x%2B1%2Bc-1
%28x%2B1%29%5E2%2Bc-1

Vertex point would be (-1,c-1).

If you want f%28-1%29=sqrt%282%29 for the minimum function value, then
c-1=sqrt%282%29
highlight%28c=1%2Bsqrt%282%29%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for vertex ( minimum in this case )
is +x%5Bv%5D+=+-b%2F%282a%29+
when the general form is:
+f%28x%29+=+a%2Ax%5E2+%2B+bx+%2B+c+
+a+=+1+
+b+=+2+
+x%5Bv%5D+=+-2%2F%282%2A1%29+
+x%5Bv%5D+=+-1+
-----------------
+f%28-1%29+=+%28-1%29%5E2+%2B+2%2A%28-1%29+%2B+c+
+f%28-1%29+=+1+-+2+%2B+c+
given:
+f%28-1%29+=+sqrt%282%29+
+sqrt%282%29+=+-1+%2B+c+
+c+=+1+%2B+sqrt%282%29+
-----------------------
Here's the plot:
+graph%28+400%2C+400%2C+-5%2C+5%2C+-5%2C+5%2C+x%5E2+%2B+2x+%2B+1+%2B+sqrt%282%29+%29+