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Graph the parabola.

y=x^(2)-10 x+27
Plot five points on the parabola: the vertex, two points to the left of the vertex button.

Graph the parabola. $\newline$ $y=x^{2}-10 x+27$ $\newline$ Plot five points on the parabola: the vertex, two points to the left of the vertex button.

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Complex conjugate theorem

Full solution

Q. Graph the parabola.

\newline

y=x^{2}-10 x+27

\newline

Plot five points on the parabola: the vertex, two points to the left of the vertex button.

Find Parabola Vertex: Find the vertex of the parabola. The vertex form of a parabola is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex. To find $h$ , use the formula $h = -b/(2a)$ . In our equation, $a = 1$ and $b = -10$ , so $h = -(-10)/(2\cdot1) = 10/2 = 5$ .
Calculate Vertex Coordinates: Find the $k$ value of the vertex by plugging $h$ into the original equation. $y = (5)^2 - 10*(5) + 27 = 25 - 50 + 27 = 2$ . So the vertex is $(5, 2)$ .
Plot Vertex on Graph: Plot the vertex $(5, 2)$ on the graph.
Calculate Left Points: Choose two $x$ -values to the left of the vertex. Let's pick $x = 3$ and $x = 4$ . Calculate the corresponding $y$ -values. For $x = 3$ : $y = (3)^2 - 10*(3) + 27 = 9 - 30 + 27 = 6$ . For $x = 4$ : $y = (4)^2 - 10*(4) + 27 = 16 - 40 + 27 = 3$ . Plot the points $(3, 6)$ and $(4, 3)$ .
Calculate Right Points: Choose two $x$ -values to the right of the vertex. Let's pick $x = 6$ and $x = 7$ . Calculate the corresponding $y$ -values. For $x = 6$ : $y = (6)^2 - 10*(6) + 27 = 36 - 60 + 27 = 3$ . For $x = 7$ : $y = (7)^2 - 10*(7) + 27 = 49 - 70 + 27 = 6$ . Plot the points $(6, 3)$ and $(7, 6)$ .

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