Question

1. solve the system of equations graphically.

y = 2(x - 1)² - 3
y = -2x + 3

1. solve the system of equations graphically.

y = -2(x - 5)² + 4
y = 2x - 6

Explanation:

Problem 4:

Step 1: Analyze the first equation \( y = 2(x - 1)^2 - 3 \)

This is a quadratic equation in vertex form \( y = a(x - h)^2 + k \), where \( a = 2 \), \( h = 1 \), \( k = -3 \). So the parabola opens upwards (since \( a > 0 \)) with vertex at \( (1, -3) \). Let's find some points:

• When \( x = 0 \), \( y = 2(0 - 1)^2 - 3 = 2(1) - 3 = -1 \)
• When \( x = 2 \), \( y = 2(2 - 1)^2 - 3 = 2(1) - 3 = -1 \)
• When \( x = 3 \), \( y = 2(3 - 1)^2 - 3 = 2(4) - 3 = 5 \)
• When \( x = -1 \), \( y = 2(-1 - 1)^2 - 3 = 2(4) - 3 = 5 \)

Step 2: Analyze the second equation \( y = -2x + 3 \)

This is a linear equation in slope - intercept form \( y=mx + b \), where \( m=-2 \) (slope) and \( b = 3 \) (y - intercept). Let's find two points:

• When \( x = 0 \), \( y=3 \)
• When \( x = 1 \), \( y=-2(1)+3 = 1 \)
• When \( x = 2 \), \( y=-2(2)+3=-1 \)

Step 3: Graph both equations and find intersection points

Plot the parabola \( y = 2(x - 1)^2 - 3 \) and the line \( y=-2x + 3 \) on the given grid. From the points we calculated, the line passes through \( (2, -1) \) which is also on the parabola (when \( x = 2 \), \( y=-1 \) for both). Let's check another intersection point. Set \( 2(x - 1)^2 - 3=-2x + 3 \)

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So \( x = 2 \) or \( x=-1 \). When \( x=-1 \), \( y=-2(-1)+3 = 5 \). And for the parabola, when \( x=-1 \), \( y = 2(-1 - 1)^2-3=5 \). So the intersection points are \( (-1,5) \) and \( (2, -1) \)

Problem 8:

Step 1: Analyze the first equation \( y=-2(x - 5)^2 + 4 \)

This is a quadratic equation in vertex form \( y = a(x - h)^2 + k \), where \( a=-2 \), \( h = 5 \), \( k = 4 \). So the parabola opens downwards (since \( a<0 \)) with vertex at \( (5,4) \). Let's find some points:

• When \( x = 4 \), \( y=-2(4 - 5)^2+4=-2(1)+4 = 2 \)
• When \( x = 6 \), \( y=-2(6 - 5)^2+4=-2(1)+4 = 2 \)
• When \( x = 3 \), \( y=-2(3 - 5)^2+4=-2(4)+4=-4 \)
• When \( x = 7 \), \( y=-2(7 - 5)^2+4=-2(4)+4=-4 \)

Step 2: Analyze the second equation \( y = 2x-6 \)

This is a linear equation in slope - intercept form \( y = mx + b \), where \( m = 2 \) (slope) and \( b=-6 \) (y - intercept). Let's find two points:

• When \( x = 3 \), \( y=2(3)-6 = 0 \)
• When \( x = 4 \), \( y=2(4)-6 = 2 \)
• When \( x = 5 \), \( y=2(5)-6 = 4 \)

Step 3: Graph both equations and find intersection points

Plot the parabola \( y=-2(x - 5)^2 + 4 \) and the line \( y = 2x-6 \) on the given grid. From the points we calculated, the line passes through \( (4,2) \) which is also on the parabola (when \( x = 4 \), \( y = 2 \) for both). Let's check another intersection point. Set \( -2(x - 5)^2+4=2x-6 \)

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So \( x = 4 \) or \( x = 5 \). When \( x = 5 \), \( y=2(5)-6 = 4 \). And for the parabola, when \( x = 5 \), \( y=-2(5 - 5)^2+4 = 4 \). So the intersection points are \( (4,2) \) and \( (5,4) \)

Problem 4 Answer:

The solutions of the system \(

$$\begin{cases}y = 2(x - 1)^2 - 3\\y=-2x + 3\end{cases}$$

\) are \( (-1,5) \) and \( (2, -1) \)

Problem 8 Answer:

The solutions of the system \(

$$\begin{cases}y=-2(x - 5)^2 + 4\\y = 2x-6\end{cases}$$

\) are \( (4,2) \) and \( (5,4) \)

Answer:

Step 1: Analyze the first equation \( y=-2(x - 5)^2 + 4 \)

This is a quadratic equation in vertex form \( y = a(x - h)^2 + k \), where \( a=-2 \), \( h = 5 \), \( k = 4 \). So the parabola opens downwards (since \( a<0 \)) with vertex at \( (5,4) \). Let's find some points:

• When \( x = 4 \), \( y=-2(4 - 5)^2+4=-2(1)+4 = 2 \)
• When \( x = 6 \), \( y=-2(6 - 5)^2+4=-2(1)+4 = 2 \)
• When \( x = 3 \), \( y=-2(3 - 5)^2+4=-2(4)+4=-4 \)
• When \( x = 7 \), \( y=-2(7 - 5)^2+4=-2(4)+4=-4 \)

Step 2: Analyze the second equation \( y = 2x-6 \)

This is a linear equation in slope - intercept form \( y = mx + b \), where \( m = 2 \) (slope) and \( b=-6 \) (y - intercept). Let's find two points:

• When \( x = 3 \), \( y=2(3)-6 = 0 \)
• When \( x = 4 \), \( y=2(4)-6 = 2 \)
• When \( x = 5 \), \( y=2(5)-6 = 4 \)

Step 3: Graph both equations and find intersection points

Plot the parabola \( y=-2(x - 5)^2 + 4 \) and the line \( y = 2x-6 \) on the given grid. From the points we calculated, the line passes through \( (4,2) \) which is also on the parabola (when \( x = 4 \), \( y = 2 \) for both). Let's check another intersection point. Set \( -2(x - 5)^2+4=2x-6 \)

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So \( x = 4 \) or \( x = 5 \). When \( x = 5 \), \( y=2(5)-6 = 4 \). And for the parabola, when \( x = 5 \), \( y=-2(5 - 5)^2+4 = 4 \). So the intersection points are \( (4,2) \) and \( (5,4) \)

Problem 4 Answer:

The solutions of the system \(

$$\begin{cases}y = 2(x - 1)^2 - 3\\y=-2x + 3\end{cases}$$

\) are \( (-1,5) \) and \( (2, -1) \)

Problem 8 Answer:

The solutions of the system \(

$$\begin{cases}y=-2(x - 5)^2 + 4\\y = 2x-6\end{cases}$$

\) are \( (4,2) \) and \( (5,4) \)