f is negative over the intervals (–∞, –3) and (7, ∞) positive over the interval ( "-3,7)," and the maximum value is 4 graph the absolute value function and equation

Answer 1

vertex has an x coordinate which is the midpoint of the interval (-3,7) = (-3+7)/2 = 4/2 =2

y = f(x) = a|x-2| + b

now find a and b

max y =4 =b

y= a|x-2| +4

a<0

(-3,0) and (7,0) are points on the graph. plot them and plot (2,4) Connect them with straight lines

line segments composing the graph are two, one upward sloping from (-3,0) to the vertex (2,4)

and a 2nd downward sloping from (2,4) to (7,0)

slopes of those lines are 4/5 and -4/5=a

f(x)=-(4/5)|x-2| +4

Answer 2

If f(x) = a|x - h| + k (graph of the absolute value function in the vertex form) is positive on the interval (-3,7) and negative otherwise, it means that it crosses x-axis at points -3 and 7. In other words these are the zeroes of the function

Also, since the function is symmetrical, the axis of symmetry should go through the midpoint of the interval (-3,7) meaning x= (-3 + 7) / 2 = 2. And this is where the maximum (vertex) is. In other words, h = 2

So the function looks like f(x) = a|x - 2| + k

f(-3) = f(7) = 0 (zeroes of the function)

f(2) = 4 (vertex)

We have the system of equations

a|7 - 2| + k = 0

a|2 - 2| + k = 4 ------------> k = 4 Substituting into the first equation we get 5a + 4 = 0 or a = -4/5

The function is f(x) = (-4/5)|x - 2| + 4