Answer:
Vertex:
[tex]f(x)=-5(x-1)^2+4[/tex]
Standard:
[tex]f(x)=-5x^2+10x-1[/tex]
Factored:
This is unfactorable.
Step-by-step explanation:
The parabola has a vertex at (1, 4) and it crosses a point at (2, -1).
We will start off with the vertex form, given by:
[tex]f(x)=a(x-h)^2+k[/tex]
Where (h, k) is the vertex.
Therefore:
[tex]f(x)=a(x-1)^2+4[/tex]
Since the function passes through (2, -1), f(x) = -1 when x = 2:
[tex]-1=a(2-1)^2+4[/tex]
Solve for a:
[tex]-5=a(1)\Rightarrow a =-5[/tex]
Therefore, vertex form is:
[tex]f(x)=-5(x-1)^2+4[/tex]
To find the standard form, expand:
[tex]f(x)=-5(x^2-2x+1)+4[/tex]
Distribute:
[tex]f(x)=-5x^2+10x-5+4[/tex]
And simplify:
[tex]f(x)=-5x^2+10x-1[/tex]
We can now factor. Which two values multiply to be 5 and add up to be 10?
Since this is no possible, the equation is unfactorable.
