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What is the quadratic function that fits the following data: (4, 9), (6, 21), (–2, –3)

Sagot :

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;

  • (4,9)
  • (6,21)
  • (-2,-3)

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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