Answer:
![f(x)=\frac{1}{2}x^2-4x+5](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B2%7Dx%5E2-4x%2B5)
Step-by-step explanation:
A parabola is written in the form
(1)
where:
is the x-coordinate of the vertex of the parabola
is the y-coordinate of the vertex of the parabola
is a scale factor
For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:
(2)
![ak=-3](https://tex.z-dn.net/?f=ak%3D-3)
From this last equation, we get that
(3)
Substituting (2) and (3) into (1) we get the new expression:
(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:
![-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6](https://tex.z-dn.net/?f=-1%3D-%5Cfrac%7B3%7D%7Bk%7D%282-4%29%5E2-3%5C%5C-1%3D-%5Cfrac%7B3%7D%7Bk%7D%5Ccdot%204%20-3%5C%5C2%3D-%5Cfrac%7B12%7D%7Bk%7D%5C%5Ck%3D-%5Cfrac%7B12%7D%7B2%7D%3D-6)
So we also get:
![a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}](https://tex.z-dn.net/?f=a%3D-%5Cfrac%7B3%7D%7Bk%7D%3D-%5Cfrac%7B3%7D%7B-6%7D%3D%5Cfrac%7B1%7D%7B2%7D)
So the equation of the parabola is:
(5)
Now we want to rewrite it in the standard form, i.e. in the form
![f(x)=ax^2+bx+c](https://tex.z-dn.net/?f=f%28x%29%3Dax%5E2%2Bbx%2Bc)
To do that, we simply rewrite (5) expliciting the various terms, we find:
![f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B2%7D%28%28x%5E2-8x%2B16%29-6%29%3D%5Cfrac%7B1%7D%7B2%7D%28x%5E2-8x%2B10%29%3D%5Cfrac%7B1%7D%7B2%7Dx%5E2-4x%2B5)