Answer:
[tex]f(x)=x^2+3x+2[/tex]
Step-by-step explanation:
We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).
Remember that the standard quadratic function is given by:
[tex]f(x)=ax^2+bx+c[/tex]
Since it passes through the point (-3, 2). This means that when [tex]x=-3[/tex], [tex]f(x)=f(-3)=2[/tex]. Hence:
[tex]f(-3)=2=a(-3)^2+b(-3)+c[/tex]
Simplify:
[tex]2=9a-3b+c[/tex]
Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:
[tex]0=a(-1)^2+b(-1)+c \\ \\0=a-b+c[/tex]
And for (1, 6):
[tex]6=a(1)^2+b(1)+c\\\\ 6=a+b+c[/tex]
So, we have a triple system of equations:
[tex]\left\{ \begin{array}{ll} 2=9a-3b+c &\\ 0=a-b+c \\6=a+b+c \end{array} \right.[/tex]
We can solve this using elimination.
Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:
[tex](0+6)=(a+a)+(-b+b)+(c+c)[/tex]
Compute:
[tex]6=2a+2c[/tex]
Let's divide both sides by 2:
[tex]3=a+c[/tex]
Now, let's eliminate b again but we will use Equation 1 and 2.
Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:
[tex]-3(0)=-3(a-b+c)[/tex]
Multiply:
[tex]0=-3a+3b-3c[/tex]
Add this to Equation 1:
[tex](0+2)=(9a-3a)+(-3b+3b)+(c-3c)[/tex]
Compute:
[tex]2=6a-2c[/tex]
Again, we can divide both sides by 2:
[tex]1=3a-c[/tex]
So, we know have two equations with only two variables:
[tex]3=a+c\text{ and } 1=3a-c[/tex]
We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:
[tex](3+1)=(a+3a)+(c-c)[/tex]
Compute:
[tex]4=4a[/tex]
Solve for a:
[tex]a=1[/tex]
So, the value of a is 1.
Using either of the two equations, we can now find c. Let's use the first one. Hence:
[tex]3=a+c[/tex]
Substitute 1 for a and solve for c:
[tex]\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}[/tex]
So, the value of c is 2.
Finally, using any of the three original equations, solve for b:
We can use Equation 3. Hence:
[tex]6=a+b+c[/tex]
Substitute in known values and solve for b:
[tex]6=(1)+b+(2)\\\\6=3+b\\\\b=3[/tex]
Therefore, a=1, b=3, and c=2.
Hence, our quadratic function is:
[tex]f(x)=x^2+3x+2[/tex]
