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Write the equation of the quadratic function given the three points on the function. Write the equation in standard form $f (x) = ax^2 +bx+ c$ . $(0, -5)$ , $(2, 27)$ , $(-2, -5)$

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

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Q. Write the equation of the quadratic function given the three points on the function. Write the equation in standard form

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

.

(0, -5)

,

(2, 27)

,

(-2, -5)

Set Up Equations: We know the standard form of a quadratic function is $f(x) = ax^2 + bx + c$ . We'll use the given points to create a system of equations to solve for $a$ , $b$ , and $c$ .
Find $c$ : Using the point $(0, -5)$ , we substitute $x = 0$ and $f(x) = -5$ into the standard form to find $c$ . $\newline$ $-5 = a(0)^2 + b(0) + c$ $\newline$ So, $c = -5$ .
Use Point $(2, 27)$ : Now we use the point $(2, 27)$ , substituting $x = 2$ and $f(x) = 27$ into the standard form. $\newline$ $27 = a(2)^2 + b(2) + c$ $\newline$ $27 = 4a + 2b - 5$ (since we already know $c = -5$ ) $\newline$ $27 + 5 = 4a + 2b$ $\newline$ $32 = 4a + 2b$
Use Point $(-2, -5)$ : Next, we use the point $(-2, -5)$ , substituting $x = -2$ and $f(x) = -5$ into the standard form. $\newline$ $-5 = a(-2)^2 + b(-2) + c$ $\newline$ $-5 = 4a - 2b + c$ $\newline$ $0 = 4a - 2b$
Solve System of Equations: We now have a system of two equations with two variables: $\newline$ $32 = 4a + 2b$ $\newline$ $0 = 4a - 2b$ $\newline$ We can solve this system by adding the two equations together to eliminate $b$ . $\newline$ $32 + 0 = 4a + 2b + 4a - 2b$ $\newline$ $32 = 8a$ $\newline$ $a = \frac{32}{8}$ $\newline$ $a = 4$
Find $b$ : With $a$ found, we can substitute it back into one of the equations to find $b$ . $0 = 4(4) - 2b$ $0 = 16 - 2b$ $-16 = -2b$ $b = \frac{-16}{-2}$ $b = 8$
Final Quadratic Function: We now have $a = 4$ , $b = 8$ , and $c = -5$ . We can write the equation of the quadratic function in standard form. $f(x) = ax^2 + bx + c$ $f(x) = 4x^2 + 8x - 5$

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