Write a formula for quadratic function if its graph has the vertex at point ( 1/3 ,−3) and passes through the point (1,1).

f(x) = 9(x -1/3)^2 -3

Step-by-step explanation:

In vertex form the equation of a quadratic with vertical scale factor "a" and vertex (h, k) is ...

y = a(x -h)^2 +k

To make the equation have (1, 1) as a solution, we need to find the value of "a". We can put the point coordinates in the equation and solve for "a":

1 = a(1 -1/3)^2 -3 . . . . . for (h, k) = (1/3, -3) as given

1 = (4/9)a -3 . . . . simplify

4 = (4/9)a . . . . . . add 3

9 = a . . . . . . . . . . multiply by 9/4

The quadratic function you desire is ...

f(x) = 9(x -1/3)^2 -3

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)

if the angular bisector of an triangle bisects the opposite side, prove that the other two sides are equal.