Omg ! the function f(x) = 4(x-7)^2-22 has been rewritten using the completing-the-square method. the function g(x) = 6x^2+12x+14 . which vertex for each function has a minimum or a maximum? what is the vertex for each function? explain your reasoning for each function

Answers:

Both functions have minima at the vertex.  The vertices are (7,-22) and (-1, 8) respectively.

Step-by-step explanation:

Hello!  Let's look at the general form for the vertex equation of a parabola:

y-k = a(x-h)^2, or (alternativelyl) y = a(x-h)^2 + k, where (h,k) is the vertex.

Comparing the given f(x) to y = a(x-h)^2 + k, we see immediately that h = 7 and k = -22, so that the vertex is (7,-22).  Since that coefficient 4 is positive, we know that the graph of this parabola opens up and that the vertex represents the minimum of f(x).

We could complete the square to rewrite g(x) in vertex form, or we could recall that the equation of the axis of symmetry is x = -b/ (2a).  Here, a = 6 and b = 12, so the axis of symmetry is x = -12 / (2*6), or x = -1.  Using synthetic division to evaluate g(x) at this x = -1, we get y = g(-1) = 8.  Thus, the vertex of g(x) is (-1, 8).  This represents the minimum value of g(x).

Both functions have minima at the vertex.

answer: i can’t tell you that

step-by-step explanation:

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.