Sure! Let's use quadratic regression to find the equation of the parabola passing through the given points [tex]\((3, -12)\)[/tex], [tex]\((-2, 43)\)[/tex], and [tex]\((2, -5)\)[/tex].
Quadratic regression aims to find a quadratic equation of the form:
[tex]\[ y = ax^2 + bx + c \][/tex]
To determine the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], we need to fit the given points to this equation.
After performing the necessary calculations, we find the coefficients:
- [tex]\(a \approx 1.0\)[/tex]
- [tex]\(b \approx -12.0\)[/tex]
- [tex]\(c \approx 15.0\)[/tex]
Thus, the quadratic equation of the parabola passing through the points [tex]\((3, -12)\)[/tex], [tex]\((-2, 43)\)[/tex], and [tex]\((2, -5)\)[/tex] is:
[tex]\[ y = 1.0 x^2 - 12.0 x + 15.0 \][/tex]
In standard form, it would simply be:
[tex]\[ y = x^2 - 12x + 15 \][/tex]
Hence, the coefficients are:
[tex]\[ \boxed{1} \][/tex] for [tex]\(x^2\)[/tex],
[tex]\[ \boxed{-12}\][/tex] for [tex]\( x \)[/tex], and
[tex]\[ \boxed{15} \][/tex] for the constant term.
