Answer:
[tex]f(x)=\frac{1}{2}x^2-4x+5[/tex]
Step-by-step explanation:
A parabola is written in the form
[tex]f(x)=a((x-h)^2+k)[/tex] (1)
where:
[tex]h[/tex] is the x-coordinate of the vertex of the parabola
[tex]ak[/tex] is the y-coordinate of the vertex of the parabola
[tex]a[/tex] is a scale factor
For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:
[tex]h=4[/tex] (2)
[tex]ak=-3[/tex]
From this last equation, we get that [tex]a=\frac{-3}{k}[/tex] (3)
Substituting (2) and (3) into (1) we get the new expression:
[tex]f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3[/tex] (4)
We also know that the parabola contains the point (2,-1), so we can substitute
x = 2
f(x) = -1
Into eq.(4) and find the value of k:
[tex]-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6[/tex]
So we also get:
[tex]a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}[/tex]
So the equation of the parabola is:
[tex]f(x)=\frac{1}{2}((x-4)^2 -6)[/tex] (5)
Now we want to rewrite it in the standard form, i.e. in the form
[tex]f(x)=ax^2+bx+c[/tex]
To do that, we simply rewrite (5) expliciting the various terms, we find:
[tex]f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5[/tex]
