f(x)=x^2+8x - 4

For this function since the a coefficient is > 1, then the parabola that the equation forms open up and its vertex is at a minimum point.

If this function is set to zero then the x intercepts for the graph can be found:

We proceed as follows: for y = 0 , Then x^{2} +8x- 4 = 0

Solving this quadratic form, x = (-b ± √ b^{2} - 4ac)/2a

a =1 , b = 8, c = - 4, by substitition x = ( -8 ± √ (8)^{2}-4(1)(-4)) / (2×1) and x = 1/2 , x = -8.5

the points (1/2 , 0) and ( -8.5,0) are the x intercepts of the graph . Knowing that the parabola opens up means that the minimum point is below the x axis . the y ordinate of this point is defenitely in the negative region of the Y axis. Since the vertex has only one x coordinate, then the only condition when the equation above has one root is when √ b2 - 4ac = 0 and x = -b/2a

x = -8/2(1) = -4 is the x coordinate of the vertex that lies in the third quadrant since the y ordinate is also negative.

To find y, substitute the value of x into the equation.

y = (-4)^{2}+ 8(-4) - 4 = -20

the minimum point is ( -4, -20) , this is the lowest point on parabola and its vertex

to test this minimum substitute x = -3 right of the vertex point , and x = -5 left of the vertex into the equation:

Y = (-3)^{2}+ 8(-3) - 4 = -19

Y = ( -5)^{2}+ 8(-5)- 4 = -19

The points on the graph are (-3, -19) , (-5 , -19) are both higher then the vertex point (-4,-20) and the point of the vertex is the minimum point the parabola openeing upward can attend.