The functions f(x) = (x + 1)2 − 2 and g(x) = -(x − 2)2 + 1 have been rewritten using the completing-the-square method. is the vertex for each function a minimum or a maximum? explain your reasoning for each function.

see explanation

Step-by-step explanation:

the equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

• If a > 0, then vertex is a minimum

• If a < 0, then the vertex is a maximum

for f(x) = (x + 1)² - 2 ← with a = 1 > 0

then vertex (- 1, - 2) is a minimum

for g(x) = - (x - 2)² + 1 ← with a = - 1 < 0

then vertex (2, 1) is a maximum

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.