Respuesta :

Answer:

[tex]\displaystyle f(x)=x^2+2x+2[/tex]

Step-by-step explanation:

System Of Linear Equations

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

[tex]\displaystyle f(x)=ax^2+bx+c[/tex]

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

[tex]\displaystyle f(-1)=a(-1)^2+b(-1)+c[/tex]

[tex]\displaystyle f(-1)=a-b+c[/tex]

[tex]\displaystyle a-b+c=1.....[eq\ 1][/tex]

Now we use the second condition f(1)=5

[tex]\displaystyle f(1)=a(1)^2+b(1)+c[/tex]

[tex]\displaystyle f(1)=a+b+c[/tex]

[tex]\displaystyle a+b+c=5.......[eq\ 2][/tex]

Finally, we use the third condition f(2)=10

[tex]\displaystyle f(2)=a(2)^2+b(2)+c[/tex]

[tex]\displaystyle f(2)=4a+2b+c[/tex]

[tex]\displaystyle 4a+2b+c=10....[eq\ 3][/tex]

We put together eq 1, eq 2, and eq 3 to form the system

[tex]\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.[/tex]

Adding the first two equations we have

[tex]\displaystyle 2a+2c=6[/tex]

[tex]\displaystyle a+c=3[/tex]

And also

[tex]\displaystyle b=2[/tex]

Using the above equation and the value of b in the third equation, we have

[tex]\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.[/tex]

Subtracting the first equation from the second

[tex]\displaystyle 3a=3[/tex]

[tex]\displaystyle a=1[/tex]

And therefore

[tex]\displaystyle c=2[/tex]

Now we have all the values, the quadratic function is

[tex]\displaystyle \boxed{f(x)=x^2+2x+2}[/tex]

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