Where is found the vertex of the function f(x) = x^2 + 5x + 1 ?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

f(x) = x^2 + 5x + 1

a = 1   b = 5    and c = 1

The vertex is V(x, y) such that:

x = -b/a = -5

y= -(b^2 - 4ab)/4a

   = -(25-4*1*1)/4 = -21/4

Then V (-5, -21/4) is located in the 3rd. quadrant.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We have to establish the quadrant where the vertex of the parable is located.

We know that the coordinates of the parabola vertex are:

 V(-b/2a;-delta/4a), where a,b,c are the coefficients of the  function and delta=b^2 -4*a*c.

y=f(x)=x^2 + 5x + 1

We'll identify the coefficients:

a=1, b=5, c=1, 2a=2, 4a=4

delta=5^2 -4*1*1=25-4=21

V(-b/2a;-delta/4a)=V(-5/2;-21/4)

Because the coordinates are both negative, the vertex is located in the third quadrant: V(-5/2;-21/4).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = x^2+5x+1.

To find the vertex.

y = x^+5x+1 =

=(x-5/2)^2  - (5/2)^2+1

 = (x-5/2)^2 - (25-4)/4 =

=(x-5/2)^2- 21/4

y+21/4 = (x-5/2)^2  which is ciparable with the standard parabola y = X^2 whose vertex is at (0,0).

So the parabola y-21/4 = (x-5/2)^2  has the  vertex V  at x =5/2 and y = -21/4. Or V(x,y) = (5/2 , -21/4).

 

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