Question

part ii: graphing systems of inequalities in two variable practice
directions: for problems #10-12, graph the system of inequalities
10.
y ≤ 2(x - 1)²
y < -x² + 5

Explanation:

Step1: Analyze \( y \leq 2(x - 1)^2 \)

This is a quadratic inequality. The parent function is \( y = x^2 \), transformed as follows: horizontal shift right 1 unit, vertical stretch by a factor of 2. The inequality \( \leq \) means we draw a solid curve (since equality is included) and shade below the curve. The vertex is at \( (1, 0) \). To find other points, plug in \( x = 0 \): \( y = 2(0 - 1)^2 = 2 \), so \( (0, 2) \) is on the curve. \( x = 2 \): \( y = 2(2 - 1)^2 = 2 \), so \( (2, 2) \) is on the curve.

Step2: Analyze \( y < -x^2 + 5 \)

This is also a quadratic inequality. The parent function \( y = x^2 \) is reflected over the x - axis and shifted up 5 units. The inequality \( < \) means we draw a dashed curve (since equality is not included) and shade below the curve. The vertex is at \( (0, 5) \). Plug in \( x = 0 \): \( y = 5 \) (vertex), \( x = 1 \): \( y=-1 + 5 = 4 \), \( x=-1 \): \( y=-1 + 5 = 4 \), \( x = 2 \): \( y=-4 + 5 = 1 \), \( x=-2 \): \( y=-4 + 5 = 1 \).

Step3: Graph the curves and shade

1. For \( y \leq 2(x - 1)^2 \): Draw a solid parabola with vertex \( (1, 0) \), passing through \( (0, 2) \) and \( (2, 2) \). Shade the region below (including the curve) this parabola.
2. For \( y < -x^2 + 5 \): Draw a dashed parabola with vertex \( (0, 5) \), passing through \( (1, 4) \), \( (-1, 4) \), \( (2, 1) \), \( (-2, 1) \). Shade the region below (not including the curve) this parabola.
3. The solution to the system is the region where the two shaded regions overlap.

Answer:

To graph the system \(

$$\begin{cases}y\leq2(x - 1)^2\\y < -x^2+5\end{cases}$$

\):

1. Graph \( y = 2(x - 1)^2 \) (solid curve):

• Vertex at \( (1, 0) \), passes through \( (0, 2) \), \( (2, 2) \). Shade below the curve (including the curve).

1. Graph \( y=-x^2 + 5 \) (dashed curve):

• Vertex at \( (0, 5) \), passes through \( (\pm1, 4) \), \( (\pm2, 1) \). Shade below the curve (excluding the curve).

1. Overlap Region: The area where both shaded regions intersect is the solution. This region is bounded below by the dashed parabola \( y=-x^2 + 5 \) and above (or within) the solid parabola \( y = 2(x - 1)^2 \), considering the respective inequality conditions.

(Note: Since this is a graphing problem, the final answer is the graphical representation as described above. To visualize, plot the two parabolas with the correct line styles (solid for \( y\leq2(x - 1)^2 \), dashed for \( y < -x^2+5 \)) and shade the overlapping region.)