Tell whether the function f(x)=1/2x^2-8x+13 has a minimum value or a maximum value. then find the minimum value or maximum value.

The minimum value of f(x) is -19.

Step-by-step explanation:

Please use parentheses around fractional coefficients to eliminate ambiguity:  f(x)=(1/2)x^2-8x+13.

Because the coefficient of the x^2 term, 1/2, is positive, the graph of this quadratic opens up.  Thus, the vertex represents the minimum of the function.

The equation of the axis of symmetry of this graph is x = -b / (2a).  In this case, x = -(-8) / (2*[1/2]), or x = 8.  Evaluating f(x) at x=8 produces the y coordinate of the vertex:  f(8) = (1/2)(8^2) - 8(8) + 13 = 32 - 64 + 13 = -19.

Thus, the vertex is (8, -19).  The minimum value of f(x) is -19.

do it yourself this is easy

step-by-step explanation:

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.