The system of equations represented by the graph in the $x y$ -plane is: $\newline$ $y=x^{2}$ $\newline$ $y=-x^{2}+2$ $\newline$ $y=x+2$

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Algebra 2

Find the vertex of the transformed function

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Q. The system of equations represented by the graph in the

x y

-plane is:

\newline

y=x^{2}

\newline

y=-x^{2}+2

\newline

y=x+2

Equations to Consider: We have three equations to consider: $\newline$ $1$ . $y = x^2$ $\newline$ $2$ . $y = -x^2 + 2$ $\newline$ $3$ . $y = x + 2$ $\newline$ To find the points of intersection, we need to set the equations equal to each other in pairs and solve for $x$ .
Intersection Points: First Two Equations: First, let's set the first two equations equal to each other to find their intersection points: $x^2 = -x^2 + 2$
Solving for x: Now, let's solve for $x$ : $2x^2 = 2$ $x^2 = 1$ $x = \pm1$
Finding y Values: We have two values for $x$ . Now we need to find the corresponding $y$ values by substituting $x$ back into one of the original equations. Let's use $y = x^2$ : $\newline$ When $x = 1$ , $y = (1)^2 = 1$ $\newline$ When $x = -1$ , $y = (-1)^2 = 1$ $\newline$ So the points of intersection between the first two equations are $(1, 1)$ and $(-1, 1)$ .
Intersection Points: Second and Third Equations: Next, let's set the second and third equations equal to each other to find their intersection points: $\newline$ $-x^2 + 2 = x + 2$
Solving for x: Now, let's solve for $x$ : $-x^2 - x = 0$ $x(x + 1) = 0$ $x = 0$ or $x = -1$
Finding y Values: We have two values for $x$ . Now we need to find the corresponding $y$ values by substituting $x$ back into one of the original equations. Let's use $y = x + 2$ : $\newline$ When $x = 0$ , $y = 0 + 2 = 2$ $\newline$ When $x = -1$ , $y = -1 + 2 = 1$ $\newline$ So the points of intersection between the second and third equations are $(0, 2)$ and $(-1, 1)$ . However, we already found $(-1, 1)$ as an intersection point between the first two equations, so we only add the new point $(0, 2)$ .
Intersection Points: First and Third Equations: Finally, let's set the first and third equations equal to each other to find their intersection points: $\newline$ $x^2 = x + 2$
Solving for x: Now, let's solve for x: $\newline$ $x^2 - x - 2 = 0$ $\newline$ $(x - 2)(x + 1) = 0$ $\newline$ $x = 2$ or $x = -1$
Finding y Values: We have two values for $x$ . Now we need to find the corresponding $y$ values by substituting $x$ back into one of the original equations. Let's use $y = x + 2$ : $\newline$ When $x = 2$ , $y = 2 + 2 = 4$ $\newline$ When $x = -1$ , $y = -1 + 2 = 1$ $\newline$ So the points of intersection between the first and third equations are $(2, 4)$ and $(-1, 1)$ . However, we already found $(-1, 1)$ as an intersection point between the first two equations and the second and third equations, so we only add the new point $(2, 4)$ .