The equation that fits into the given data is
[tex]x^2+2x-6=0 \\[/tex]
Quadratic Equation
This is an algebraic equation or expression that have a second degree as it highest power. An example of quadratic equation is given below
[tex]f(x)=ax^2+bx+c\\[/tex]
The given data are set of points attached to a quadratic equation and we can use it to find the quadratic equation.
given that [tex]f(x)=ax^2+bx+c\\[/tex]
Data;
The point at (4)(9) will give us the equation
Let's substitute the value of x and y in the equation and solve
[tex]f(4)=9\\a(4)^2+b(4)+c=9\\16a^2+4b+c-0...equation 1[/tex]
The point at(6)(21) will give the equation
Let's substitute the value of x and y in the equation and solve
[tex]f(6)=21\\a(6)^2+b(6)+c=21\\36a+6b+c=21...equation 2[/tex]
The point at (-2)(-3) gives the equation
Let's substitute the value of x and y in the equation and solve
[tex]f(-2)=-3\\a(-2)^2+b(-2)+c=-3\\4a-2b+c=-3...equation 3[/tex]
We have the following system of equation
[tex]16a+4b+c=9...equation1\\36a+6b+c=21...equation2\\4a-2b+c=-3...equation3\\[/tex]
When we solve the above system of equation either using elimination or substitution method, we have
[tex]a=1/2, b=1, c= -3[/tex]
substituting these values into f(x)
[tex](\frac{1}{2})x^2+(1) + (-3)=0\\\frac{1}{2}x^2+x-3=0\\[/tex]
multiply through by 2
[tex](2)\frac{1}{2}x^2+(2)x-(2)3=0\\x^2+2x-6=0[/tex]
The quadratic equation is [tex]x^2+2x-6=0[/tex]
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