remark: the x-intercepts are clearly (0, 0) and (20, 0), not (0, 0), (0, 20).
having the 2 x intercepts and the vertex we can draw the graph, check picture 1 attached.
Moreover, we can also determine the quadratic function as follows.
The quadratic function with roots a and b, is given by the equation:
f(x)=c(x-a)(x-b).
[ since f(a)=0, f(b)=0, f(x) must have one factor (x-a) and one (x-b)]
The x coordinates of the x intercepts are the roots of the function so with the roots given:
f(x)=c(x-0)(x-20)=cx(x-20), the vertex is at (10, 15) so
15=f(10)=c*10*(10-20)=c*10*(-10)=-100c
c=-15/100=-3/20
the quadratic function (expressed in factorized form) is
f(x)=-3/20x(x-20)
part A: the x intercept 0 is the point where the ball was hit, the x intercept 20 is the point where the ball fell back to the ground, 20 feet away from the kicker.
the function is increasing in the interval x∈(0, 10) and decreasing in x∈(10, 20)
this means that the height is increasing in the interval (0, 10) and decreasing as x goes through the interval (10,20)
the distance from the kicker is increasing during the whole interval (0, 20)
Part B: at x=15, f(x)=f(15)=-3/20*15(15-20)=[tex] \frac{-3}{20}*15*(-5)= \frac{-3}{4}*3*(-5)= \frac{45}{4} [/tex]
at x=12, f(x)=f(12)=-3/20*12(12-20)=[tex] \frac{-3}{20}*12*(-8)= \frac{-3}{5}*3*(-8)= \frac{72}{5} [/tex]
the average rate of change is the ratio of the differences in y to the differences in x : [tex] \frac{y_1_5-y_1_2}{x_1_5-x_1_2}= \frac{\frac{45}{4}-\frac{72}{5}}{15-12}= -1.05[/tex]
so the average rate of change is -1.05. The minus, tells us that the height is decreasing in the following way: for 1 feet in distance, we are loosing 1.05 feet in height.
