Three points on the graph of the function $f(x)$ are $(0,4)$ $(1,5)$ and $(2,8)$ which represents $f(x)$

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Algebra 2

Find the vertex of a parabola

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Q. Three points on the graph of the function

f(x)

are

(0,4)

(1,5)

and

(2,8)

which represents

f(x)

Identify Function Form: Determine the general form of the function. $\newline$ Since we have three points, we can assume that the function $f(x)$ is a quadratic function of the form $f(x) = ax^2 + bx + c$ .
Set Up Equations: Set up a system of equations using the given points. $\newline$ Substitute the $x$ and $y$ values from the points into the quadratic equation to create three equations. $\newline$ For point $(0,4)$ : $\newline$ $4 = a(0)^2 + b(0) + c$ $\newline$ $4 = c$ $\newline$ For point $(1,5)$ : $\newline$ $5 = a(1)^2 + b(1) + c$ $\newline$ $5 = a + b + c$ $\newline$ For point $(2,8)$ : $\newline$ $8 = a(2)^2 + b(2) + c$ $\newline$ $y$ $0$
Solve Equations: Solve the system of equations. $\newline$ We already know that $c = 4$ from the first equation. Now we can substitute $c$ into the other two equations. $\newline$ Substitute $c = 4$ into the second equation: $\newline$ $5 = a + b + 4$ $\newline$ $a + b = 1$ $\newline$ Substitute $c = 4$ into the third equation: $\newline$ $8 = 4a + 2b + 4$ $\newline$ $4a + 2b = 4$
Find $a$ and $b$ : Solve for $a$ and $b$ . We have two equations now: $a + b = 1$ $4a + 2b = 4$ We can multiply the first equation by $2$ to help eliminate $b$ : $2a + 2b = 2$ Now subtract this new equation from the second equation: $(4a + 2b) - (2a + 2b) = 4 - 2$ $2a = 2$ $a = 1$ Now substitute $a = 1$ into the first equation: $1 + b = 1$ $b = 0$
Write Final Function: Write the final function. $\newline$ Now that we have $a = 1$ , $b = 0$ , and $c = 4$ , we can write the function $f(x)$ : $\newline$ $f(x) = 1x^2 + 0x + 4$ $\newline$ $f(x) = x^2 + 4$