Answer:
7(b+5) is the average rate of change of the line y=7x^2-4 between points
Explanation:
We'll assume the equation reads f(x)=7x^2 - 4 and not f(x)=7x2−4.
It is unclear what the value of b should be in the interval [5,b]. Lets assume b is 6, which will provide a second point on the hyperbola that is reasonably close to the first, at 5. Calculate f(x) for x = 5 and 6. The two points are (5, 171) and (6,248).
The slope of the line that connects the points (5, 171) and (6,248) and be calculated by the Rise/Run for the two points. Rise = 77, Run = 1, for a slope of (Rise/Run) 77/1 or 77. The y intersect, b, can then be calculated from the resulting y= 77x + b by entering one of the two known points and solving for b. [Note: this is not the same b in the interval [5,b]. The slope of 77 is the average slope for the interval [5,6].
We can use the same procedure for the interval [5,b]. The first point, (5,171) is the same. The second point (b, f(b)) has b as an unknown.
The slope for this line is:
Point 1: (5, 171)
Point 2: (b,f(b))
where f(b) = 7b^2-4
Point 2: (b, 7b^2-4)
The slope is:
Rise: (7b^2-4) - 171
Rise: 7b^2 - 175
Run: b - 5
The slope is Rise/Run or ((7b^2- 175)/(b-5)
Lets factor the numerator: (7b^2- 175) = 7*(b^2 - 25)
7*(b^2 - 25) = 7*(b+5)(b-5)
(7*(b+5)(b-5))/(b-5)
7(b+5)
This is the slope, or average rate of change, between points 5 and b on the line f(x) = 7x^2 -4,