[tex]\boxed{f(x)=-\frac{3}{2}x^2+3x+\frac{9}{2}}[/tex]
The Standard Form of the equation of a parabola is given by the form:
[tex]y=ax^2+bx+c[/tex]
As y is a function of x, then we can write:
[tex]y=f(x)=ax^2+bx+c \\ \\ a, \ b, c \ Real \ Coefficients[/tex]
Also, we can write [tex]f(x)[/tex] in factored form as:
[tex]f(x)=a(x-x_{1})(x-(x_{2}) \\ \\ x_{1} \ and \ x_{2} \ Are \ the \ roots[/tex]
In this case:
[tex]x_{1}=-1 \\ \\ x_{2}=3[/tex]
So:
[tex]f(x)=a(x-(-1))(x-3) \\ \\ f(x)=a(x+1)(x-3)[/tex]
We also know that the parabola passes through the point (1,6):
[tex]f(1)=6=a(1+1)(1-3) \\ \\ Isolating \ a: \\ \\ 6=a(2)(-2) \\ \\ 6=a(-4) \\ \\ a=-\frac{6}{4} \\ \\ a=-\frac{3}{2}[/tex]
Finally, the equation of the parabola is:
[tex]f(x)=-\frac{3}{2}(x+1)(x-3)[/tex]
Or applying Distributive Property:
[tex]f(x)=-\frac{3}{2}\left(x^2-2x-3\right) \\ \\ \boxed{f(x)=-\frac{3}{2}x^2+3x+\frac{9}{2}}[/tex]
Piecewise-defined function: https://brainly.com/question/12169258
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