Where is found the vertex of the function f(x) = x^2 + 5x + 1 ?
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.
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
Because the coordinates are both negative, the vertex is located in the third quadrant: V(-5/2;-21/4).
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 =
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).