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Write the equation in standard form $f(x) = ax^2 + bx + c$ . $(0, -12)$ . $(1, -9)$ , $(-2, -36)$

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

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Q. Write the equation in standard form

f(x) = ax^2 + bx + c

.

(0, -12)

.

(1, -9)

,

(-2, -36)

Create Equations: We have three points: $(0, -12)$ , $(1, -9)$ , and $(-2, -36)$ . Let's use these points to create a system of equations based on the standard form of a parabola. $f(x) = ax^2 + bx + c$ Using the points to create equations: For $(0, -12)$ : $-12 = a(0)^2 + b(0) + c$ For $(1, -9)$ : $-9 = a(1)^2 + b(1) + c$ For $(-2, -36)$ : $-36 = a(-2)^2 + b(-2) + c$
Simplify Equations: First, let's simplify the equations: $\newline$ For $(0, -12)$ : $c = -12$ $\newline$ For $(1, -9)$ : $a + b + c = -9$ $\newline$ For $(-2, -36)$ : $4a - 2b + c = -36$
Substitute $c$ : Now we know that $c = -12$ , we can substitute it into the other two equations: $\newline$ $a + b - 12 = -9$ $\newline$ $4a - 2b - 12 = -36$
Eliminate b: Let's simplify the equations further: $\newline$ $a + b = -9 + 12$ $\newline$ $a + b = 3$ $\newline$ $4a - 2b = -36 + 12$ $\newline$ $4a - 2b = -24$
Solve for $a$ : Now we have a system of two equations with two variables: $\newline$ $a + b = 3$ $\newline$ $4a - 2b = -24$ $\newline$ Let's multiply the first equation by $2$ to help us eliminate $b$ : $\newline$ $2a + 2b = 6$ $\newline$ $4a - 2b = -24$
Find $b$ : Add the two equations together to eliminate $b$ :
$(2a + 2b) + (4a - 2b) = 6 + (-24)$
$6a = -18$
Divide both sides by $6$ to solve for $a$ :
$a = \frac{-18}{6}$
$a = -3$
Find $b$ : Add the two equations together to eliminate $b$ :
$(2a + 2b) + (4a - 2b) = 6 + (-24)$
$6a = -18$
Divide both sides by $6$ to solve for $a$ :
$a = -18 / 6$
$a = -3$ Now that we have $a$ , let's substitute it back into one of the original equations to find $b$ :
$b$ $0$
$b$ $1$
$b$ $2$
$b$ $3$

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