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$y=x^2-6(x-1)$ Which of the following is an equivalent form of the equation of the graph shown in the $xy$ -plane, from which the coordinates of vertex $V$ can be identified as constants in the equation?

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Convert equations of conic sections from general to standard form

Full solution

Q.

y=x^2-6(x-1)

Which of the following is an equivalent form of the equation of the graph shown in the

xy

-plane, from which the coordinates of vertex

V

can be identified as constants in the equation?

Expand Equation: Expand the equation $y = x^2 - 6(x - 1)$ to its standard form. $\newline$ We distribute the $-6$ to both terms inside the parentheses. $\newline$ $y = x^2 - 6x + 6$
Convert to Vertex Form: Convert the standard form of the equation into vertex form. $\newline$ The vertex form of a quadratic equation is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. $\newline$ To convert the standard form $y = x^2 - 6x + 6$ into vertex form, we need to complete the square. $\newline$ First, we factor out the coefficient of $x^2$ , which is $1$ in this case, so we can leave the equation as is. $\newline$ Next, we take the coefficient of $x$ , which is $-6$ , divide it by $2$ , and square it to find the value to complete the square. $\newline$ $(-6/2)^2 = (-3)^2 = 9$ $\newline$ We add and subtract this value inside the equation to complete the square. $\newline$ $y = x^2 - 6x + 9 - 9 + 6$
Rewrite with Completed Square: Rewrite the equation with the completed square and simplify. $\newline$ $y = (x^2 - 6x + 9) - 3$ $\newline$ Now, we can rewrite the equation as a perfect square trinomial. $\newline$ $y = (x - 3)^2 - 3$ $\newline$ This is the vertex form of the equation, and from this form, we can identify the vertex $V$ of the parabola. $\newline$ The vertex $V$ has coordinates $(h, k) = (3, -3)$ .

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