How to Find the Vertex of f(x)=x^2+4x+1 | Step-by-Step Guide

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In summary, the function f(x)=x^2+4x+1 has a vertex at the point (-2,-3) and the x-coordinate of the vertex is -2. This can be found by using the Completing the Square method or by finding the turning point of the function.
  • #1
poohbear1986
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Consider the function f(x)=x^2+4x+1

a)Find h, the x-coordinate of the vertex of this parabola.
 
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  • #2
poohbear1986 said:
Consider the function f(x)=x^2+4x+1

a)Find h, the x-coordinate of the vertex of this parabola.

This must be related to Completing the Square. Your new version of the function (standard form) will contain information for the translated (moved away from standard position) graph, based on (h, k); you read this point directly from the function in standard form. Your book explains Completing the Square; are you using no intermediate algebra or no "elementary functions" book?
 
  • #3
How can you possibly "make sure you are doing this right" when you haven't done anything? Show us what you have done and then we can tell you whether you are doing it right or not.
 
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  • #4
f(x) = x^2 + 4x + 1
= x^2 + 4x + (2)^2 - (2)^2 + 1
= (x + 2)^2 - 3

coordinates of the turning point (-2, -3), hence the x-coordinate, h = -2

OR

f(x) = x^2 + 4x + 1

f '(x) = 2x + 4

at the turning point f '(x) = 0

2x + 4 = 0

2x = -4

x = -2

therefore, h the x-coordinate of the vertex of this parabola equals to -2

Hope my answer is correct...
 

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