Ty Use the graph to find the local minimum and the local
maximum for the given function.
Over the interval (-3, 0], the local minimum is
Over the interval [0, 3), the local maximum is
v
Over the interval [0, 3), the local minimum is
v
-41
18
412
-16
) Intro
Done
Type here to search
0
it

Which quadratic function best fits this data? x y 1 32 2 78 3 178 4 326 5 390 6 337 a. y=11.41x2+154.42x−143.9 b. y=−11.41x2+154.42x−143.9 c. y=11.41x2+154.42x+143.9 d. y=−11.41x2+154.42x+143.9

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.

Find the value of x. round the length to the nearest tenth