SOLUTION: graph the quadratic function f(x)=(x+3)^2+1

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Question 407300: graph the quadratic function f(x)=(x+3)^2+1
Found 2 solutions by MathLover1, graphmatics:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=%28x%2B3%29%5E2%2B1
f%28x%29=%28x%5E2+%2B6x+%2B9%29%2B1

f%28x%29=x%5E2+%2B6x+%2B10
use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ...you have a=1, b=6 and c=10

x+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A10+%29%29%2F%282%2A1%29+


x+=+%28-6+%2B-+sqrt%28+36-40+%29%29%2F2+

x+=+%28-6+%2B-+sqrt%28+-4+%29%29%2F2+
x+=+%28-6+%2B-+sqrt%28+4i+%29%29%2F2+

x+=+%28-6+%2B-+%28+2i+%29%29%2F2+

x1+=+%28-6+%2B%28+2i+%29%29%2F2+
x1+=+%28-3+%2B%28+i+%29%29
x1+=+%28-3+-%28+i+%29%29
graph:
+graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+x%5E2+%2B6x+%2B10%29+





Answer by graphmatics(170) About Me  (Show Source):
You can put this solution on YOUR website!
Put y for f(x) and get
y = (x+3)^2 +1
y = x^2 + 6*x +9 +1
y = x^2 + 6*x + 10
we plot a group of quadratic equation (x,y) points for a group of x say {0, 1,-1,2,-2,3,-3}.
if x=0 then y = x^2 + 6*0 +10 so y=10 and (0,10) is a solution point for the quadratic equation y = x^2 + 6*x + 10.
For x=-3 y=3*2+6*3+10=17 and (-3,1) is a point and similarly for
x=-1,x=2,x=-2,x=3,x=-3
+graph%28+500%2C+500%2C+-6%2C+5%2C+-10%2C+10%2C+0%2C+x%5E2%2B6%2Ax%2B10%29+