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.

step-by-step explanation: h: (- 3, 6)

m: (1, - 2)

o: (7, 6)

r: (- 9, - 2)

midpoint of hm: (- 1, 2)

midpoint of or: (- 1, 2)

therefore, b is the midpoint of both diagonals bisect each other !

to prove the diagonals are perpendicular, show that the product of the slopes = - 1

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead