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77julia77 [94]
3 years ago
13

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Mathematics
1 answer:
Serjik [45]3 years ago
6 0

Answer:

\displaystyle f(x)=x^2+2x+2

Step-by-step explanation:

<u>System Of Linear Equations </u>

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

\displaystyle f(x)=ax^2+bx+c

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

\displaystyle f(-1)=a(-1)^2+b(-1)+c

\displaystyle f(-1)=a-b+c

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

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

\displaystyle f(1)=a(1)^2+b(1)+c

\displaystyle f(1)=a+b+c

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

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

\displaystyle f(2)=a(2)^2+b(2)+c

\displaystyle f(2)=4a+2b+c

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

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

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

Adding the first two equations we have

\displaystyle 2a+2c=6

\displaystyle a+c=3

And also

\displaystyle b=2

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

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

Subtracting the first equation from the second

\displaystyle 3a=3

\displaystyle a=1

And therefore

\displaystyle c=2

Now we have all the values, the quadratic function is

\displaystyle \boxed{f(x)=x^2+2x+2}

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