SOLUTION: I am lost I'm not sure what this is asking? To plot my points help. my brain is mush right now. Consider the function f(x)= x^2+4x+1 Find h, the x-coordinate of the ve

Algebra ->  Functions -> SOLUTION: I am lost I'm not sure what this is asking? To plot my points help. my brain is mush right now. Consider the function f(x)= x^2+4x+1 Find h, the x-coordinate of the ve      Log On


   

Question 92007: I am lost
I'm not sure what this is asking? To plot my points help. my brain is mush right now.
Consider the function f(x)= x^2+4x+1
Find h, the x-coordinate of the vertex of this parabola.

Substitute the two whole number values immediately to the left and right of h into the function to find the corresponding y. Fill in the following table. Make sure your x-values are in increasing order in your table.
x __|_y_
____|___
____|___
h=__|___
____|___
____|___
Answer by mathispowerful(115) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+x%5E2%2B4x%2B1
Let's find the vertex first
in standard format of quadratic equation y+=+ax%5E2%2Bbx%2Bc
the x value of vertex is given by -b%2F%282a%29
in your function, a=1, b =4, so
x = -4%2F%282%2A1%29 = -2
then f(-2) = -3
so vertex is (-2, -3)
Second part:
X ---------Y

-4 -------- 1
-3 -------- -2
-2 -------- -3
-1 -------- -2
0 -------- 1
from these numbers, you can see that they are symmetric, that's the characteristic of a parabola