Answer:
[tex]\large\boxed{\text{Factored Form:}\ f(x)+-(x-1)(x-5)}\\\boxed{\text{Vertex Form:}\ f(x)=-(x-3)^2+4}\\\boxed{\text{Standard Form:}\ f(x)=-x^2+6x-5}[/tex]
Step-by-step explanation:
(look at the picture)
Factored form:
[tex]f(x)=a(x-x_1)(x-x_2)[/tex]
x₁, x₂ - zeros
Vertex form:
[tex]f(x)=a(x-h)^2+k[/tex]
(h, k) - vertex
Standard form:
[tex]f(x)=ax^2+bx+c[/tex]
If from the vertex we go 1 unit down (up) and 1 unit left (right) and we get the point on the parabola, then a = 1.
The parabola is open down, therefore a < 0 → a = -1.
The zeros are [tex]x_1=1[/tex] and [tex]x_2=5[/tex]. Therefore the Factored Form is:
[tex]f(x)=-(x-1)(x-5)[/tex]
The vertex is V(3, 4). Therefore the vertex form is:
[tex]f(x)=-(x-3)^2+4[/tex]
Convert it to a standard form using (a - b)² = a² - 2ab + b²
[tex]f(x)=-(x^2-2(x)(3)+3^2)+4=-x^2+6x-9+4=-x^2+6x-5[/tex]
