SOLUTION: Give the vertex and 4 other points on the graph. y=x^2+2x+6

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Question 1038648: Give the vertex and 4 other points on the graph. y=x^2+2x+6
Answer by PRMath(133) About Me  (Show Source):
You can put this solution on YOUR website!
y = x^2 + 2x + 6
To find the vertex of an equation, you just have to use this formula:
-b%2F2a
In this format, the following values can be seen:
a = 1
b = 2
c = 6
SO!
-2%2F2%281%29
That gives us:
-1
NOW you know that -1 is the x value of the vertex. We have to find the y value, so we just "plug in" the -1 into the "X" values in the original equation. Let's do that now:
y = x^2 + 2x + 6
y = (-1)^2 +2(-1) +6
y = 1 -2 +6
y = 5
SO! Now we have an x value of -1 and a y value of 5.
That is our vertex: (-1, 5)


To find other points, you can just set up some x values and determine the y.
We already know that y is 5 when x is -1. Let's do a few more....

y = x^2 + 2x + 6

x ---------------- y
-2 --------------- (-2)^2 + 2(-2) + 6 = 6 (-2, 6)
0 ----------------- (0)^2 +2(0) + 6 = 6 (0, 6)
-4 ---------------- (-4)^2 +2(-4) + 6 = 14 (-4, 14)
2 ------------------(2)^2 +2(2) + 6 = 14 (2, 14)

Those are four more points on this graph.
I hope this helps!