Which function represents g(x), a reflection of f(x) = on a coordinate plane, 2 exponential functions are shown. g (x) decreases in quadrant 2 and approaches y = 0 in quadrant 1. it goes through (negative 1, 1) and crosses the y - axis at (0, 0.)x across the y-axis? g(x) = 2(3)x g(x) = −one-half(3)x g(x) = one-half(3)−x g(x) = 2(3)−x

Step-by-step explanation:

Given:

The graph of function are given.

The equation for is given as:

Now, the graph of is a reflection of .

The graph of passes through the point (-1, 1.5) [second quadrant] and crosses the y-axis at (0, 0.5).

As evident from the graph, the functions are reflections about the y-axis.

We know the transformation rule for reflection about the y-axis as:

Now, the y-intercept of the function is obtained by plugging in . This gives,

So, the y-intercept is at (0, 1). But the graph of crosses the y-axis at (0, 0.5). As we observe, the coordinate rule for the transformation can be written as:

(0, 1) → (0, 0.5)

So, the reflected graph is compressed vertically by a factor of .

Therefore, the transformation is given as:

Therefore, the equation for is:

6(h+1)/2=a

a=area

step-by-step explanation:

its the same thing as a square or rectangle but after multiplying two sides you need to divide it by two.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.