Answer:
Part a) The function’s equation written in vertex form is
![f(x)=2(x-2)^2-8](https://tex.z-dn.net/?f=f%28x%29%3D2%28x-2%29%5E2-8)
Part b) The function’s equation written in factored form is equal to
![f(x)=2x(x-4)](https://tex.z-dn.net/?f=f%28x%29%3D2x%28x-4%29)
Step-by-step explanation:
Part a) What is the function’s equation written in vertex form?
we know that
The equation of a vertical parabola written in vertex form is equal to
![f(x)=a(x-h)^2+k](https://tex.z-dn.net/?f=f%28x%29%3Da%28x-h%29%5E2%2Bk)
where
a is a coefficient
(h,k) is the vertex
Looking at the graph
The vertex is the point (2,-8)
substitute
![f(x)=a(x-2)^2-8](https://tex.z-dn.net/?f=f%28x%29%3Da%28x-2%29%5E2-8)
Find the value of the coefficient a
take one point from the graph
(0,0)
substitute in the equation
![0=a(0-2)^2-8\\0=4a-8\\4a=8\\a=2](https://tex.z-dn.net/?f=0%3Da%280-2%29%5E2-8%5C%5C0%3D4a-8%5C%5C4a%3D8%5C%5Ca%3D2)
therefore
The function’s equation written in vertex form is
![f(x)=2(x-2)^2-8](https://tex.z-dn.net/?f=f%28x%29%3D2%28x-2%29%5E2-8)
Part b) What is the function’s equation written in factored form?
we know that
The equation of a vertical parabola written in factored form is equal to
![f(x)=a(x-x_1)(x-x_2)](https://tex.z-dn.net/?f=f%28x%29%3Da%28x-x_1%29%28x-x_2%29)
where
a is a coefficient
x_1 and x_2 are the zeros or x-intercepts of the function
Remember that the x-intercept is the value of x when the value of the function is equal to zero
Looking at the graph
The zeros or x-intercepts of the function are
x=0 and x=4
so
![f(x)=a(x-0)(x-4)](https://tex.z-dn.net/?f=f%28x%29%3Da%28x-0%29%28x-4%29)
![f(x)=ax(x-4)](https://tex.z-dn.net/?f=f%28x%29%3Dax%28x-4%29)
Find the value of the coefficient a
take one point from the graph
(2,-8)
substitute
![-8=a(2)(2-4)\\-8=-4a\\a=2](https://tex.z-dn.net/?f=-8%3Da%282%29%282-4%29%5C%5C-8%3D-4a%5C%5Ca%3D2)
therefore
The function’s equation written in factored form is equal to
![f(x)=2x(x-4)](https://tex.z-dn.net/?f=f%28x%29%3D2x%28x-4%29)