# Want to make sure im doing this right

1. Nov 2, 2007

### poohbear1986

Consider the function f(x)=x^2+4x+1

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

2. Nov 2, 2007

### symbolipoint

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. Nov 3, 2007

### HallsofIvy

Staff Emeritus
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.

Last edited: Nov 3, 2007
4. Nov 4, 2007

### King Mickey

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