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Write a quadratic equation for a parabola that passes through the points $(-1,-3)$ , $(0,-4)$ , and $(2,6)$ using quadratic regression.

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Write a quadratic function from its x-intercepts and another point

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Q. Write a quadratic equation for a parabola that passes through the points

(-1,-3)

,

(0,-4)

, and

(2,6)

using quadratic regression.

Set up equations: We need to find the quadratic equation in the form $y = ax^2 + bx + c$ . We'll use the given points to set up a system of equations. $\newline$ Using the points $(-1,-3)$ , $(0,-4)$ , and $(2,6)$ , we substitute into $y = ax^2 + bx + c$ : $\newline$ For $(-1,-3)$ : $-3 = a(-1)^2 + b(-1) + c$ $\newline$ For $(0,-4)$ : $-4 = a(0)^2 + b(0) + c$ $\newline$ For $(2,6)$ : $(-1,-3)$ $0$
Simplify system: Simplify and write the system of equations: $\newline$ $1$ . $-3 = a - b + c$ $\newline$ $2$ . $-4 = c$ $\newline$ $3$ . $6 = 4a + 2b + c$
Substitute and simplify: Substitute $c = -4$ into equations $1$ and $3$ : $\newline$ $1.$ $-3 = a - b - 4$ $\newline$ $3.$ $6 = 4a + 2b - 4$
Solve for variables: Simplify the equations: $\newline$ $1$ . $1 = a - b$ $\newline$ $3$ . $10 = 4a + 2b$
Find $b$ value: Solve the system using substitution or elimination. From equation $1$ , $a = b + 1$ . Substitute into equation $3$ : $\newline$ $10 = 4(b + 1) + 2b$ $\newline$ $10 = 4b + 4 + 2b$ $\newline$ $10 = 6b + 4$
Find a value: Solve for $b$ : $6 = 6b$ $b = 1$
Write final equation: Substitute $b = 1$ into $a = b + 1$ : $a = 1 + 1$ $a = 2$
Write final equation: Substitute $b = 1$ into $a = b + 1$ :
$a = 1 + 1$
$a = 2$ Now we have $a = 2$ , $b = 1$ , and $c = -4$ . Write the final quadratic equation:
$y = 2x^2 + x - 4$

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