what is the minimum value of f(x)=x^2+4x-8

Question

where does the min or max occur and what is the value

Answer 1

The standard form of a quadratic equation is:

 

y = ax² + bx + c

 

where a, b, and c are constants.  If the constant "a" is > 0, the the parabola opens upward and the vertex is a minimum.  If "a" <0, the parabola is inverted and the vertex is the maximum.  Your equation is:

 

f(x) = x² + 4x - 8

 

What are the values of a, b, and c?  Is a>0 or < 0?  Is the vertex a maximum or minimum? To find where the vertex occurs along the x-axis, use the formula:

 

x = -b/(2a)

 

To find the value of the vertex, plug the x-coordinate of the vertex (-b/2a) into f(x) and compute the value.

Answer 2

The graph of y = Ax²+Bx+C where A > 0 is a parabola opening upward.  The lowest point on the graph is the vertex with x coordinate -B/(2A).

 

So, for f(x) = x² + 4x - 8, the lowest point has x-coordinate x = -4/(2(1)) = -2

 

   Minimum value = f(-2) = -12

 

Note:  If A<0, the graph has a highest point, so there would be a maximum value in that case.