Find the equation of the parabola with points (-3,15), (0,-6), & (2,10)

The equation of a parabola through thee points can be found different ways. One is to use a quadratic regression tool. Another is to write and solve linear equations in the coefficients. (3 equations in 3 unknowns).

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Here, we can eliminate answer choices that don't work to arrive at the correct answer choice.

The point (0, -6) is the y-intercept of the function. That means the value of the constant term in the quadratic is -6. (Eliminates A and C.)

The y-values of the other two points are both greater than -6, indicating the parabola opens upward. That means the leading coefficient is positive. (Eliminates B.)

The only reasonable choice is D:

y = 3x² +2x -6

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Additional comment

You get the same answer if you use a regression tool.

jacob193 jacob193 · Answer 2 · 2022-03-10T23:29:06+01:00

Answer:

[tex]y = 3\, x^{2} + 2\, x - 6[/tex].

Step-by-step explanation:

In general, the equation of a parabola is in the form [tex]y = a\, x^{2} + b\, x + c[/tex] for some constants [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], where [tex]a \ne 0[/tex].

Let [tex]y = a\, x^{2} + b\, x + c\![/tex] denote the equation of this parabola for some constants [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] where [tex]a \ne 0[/tex]. A point [tex](x_{0},\, y_{0})[/tex] is on this parabola if and only if the equation of this parabola holds after substituting in [tex]x = x_{0}[/tex] and [tex]y = y_{0}[/tex]:

[tex]y_{0} = a\, {x_{0}}^{2} + b\, x_{0} + c[/tex].

Thus, each of the three distinct points on this parabola would give a equation about [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex]:

The equation for [tex](-3,\, 15)[/tex] would be [tex]15 = (-3)^{2}\, a + (-3)\, b + c[/tex].
The equation for [tex](0,\, 6)[/tex] would be [tex]6 = 0^{2}\, a + 0\, b + c[/tex].
The equation for [tex](2,\, 10)[/tex] would be [tex]10 = 2^{2}\, a + 2\, b + c[/tex].

Simplify the equations:

[tex]\left\lbrace\begin{aligned}& 9\, a - 3\, b + c = 15 \\ & c = 6 \\ & 4\, a + 2\, b + c = 10\end{aligned}\right.[/tex].

Solve this linear system of three equations and three unknowns for [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex]:

[tex]\left\lbrace \begin{aligned} & a = 3 \\ & b = 2 \\ & c = (-6) \end{aligned}\right.[/tex].

Therefore, the equation of this parabola would be:

[tex]y = 3\, x^{2} + 2\, x - 6[/tex].