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Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an
(x,y) point.

y=-x^(2)+10 x-26
Answer:

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an $(x, y)$ point. $\newline$ $y=-x^{2}+10 x-26$ $\newline$ Answer:

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Solve exponential equations by rewriting the base

Full solution

Q. Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an

(x, y)

point.

\newline

y=-x^{2}+10 x-26

\newline

Answer:

Identify Quadratic Equation: Identify the quadratic equation in standard form. $\newline$ The given quadratic equation is $y = -x^2 + 10x - 26$ . This is already in the standard form $y = ax^2 + bx + c$ , where $a = -1$ , $b = 10$ , and $c = -26$ .
Use Vertex Formula: Use the vertex formula to find the $x$ -coordinate of the vertex. $\newline$ The $x$ -coordinate of the vertex of a parabola in the form $y = ax^2 + bx + c$ is given by $-\frac{b}{2a}$ . Here, $a = -1$ and $b = 10$ . $\newline$ So, $x = -\frac{10}{2 \times -1} = -\frac{10}{-2} = 5$ .
Substitute Coordinates: Substitute the $x$ -coordinate back into the original equation to find the $y$ -coordinate of the vertex. $\newline$ Now that we have $x = 5$ , we substitute it back into the equation $y = -x^2 + 10x - 26$ to find the $y$ -coordinate. $\newline$ $y = -(5)^2 + 10(5) - 26$ $\newline$ $y = -25 + 50 - 26$ $\newline$ $y = 25 - 26$ $\newline$ $y = -1$ .

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