Question

graph each quadratic inequality on the coordinate plane provided to display the solution set.

1. ( y geq x^2 + 4x - 5 )
2. ( y > -x^2 + 4x - 3 )
3. ( y leq -3(x + 1)^2 + 10 )

Explanation:

Problem 1: \( y \geq x^2 + 4x - 5 \)

Step 1: Rewrite the quadratic in vertex form (or find roots)

First, complete the square for \( x^2 + 4x - 5 \).
\( x^2 + 4x - 5 = (x^2 + 4x + 4) - 4 - 5 = (x + 2)^2 - 9 \).
So the parabola has vertex at \( (-2, -9) \), and since the coefficient of \( x^2 \) is positive (\( a = 1 > 0 \)), it opens upward.

Step 2: Determine the boundary line

The inequality is \( y \geq (x + 2)^2 - 9 \), so the boundary is the parabola \( y = (x + 2)^2 - 9 \) (solid line, because \( \geq \) includes equality).

Step 3: Test a point to shade the region

Test the origin \( (0, 0) \):
\( 0 \geq (0 + 2)^2 - 9 \)?
\( 0 \geq 4 - 9 \)?
\( 0 \geq -5 \), which is true. So shade the region above (including) the parabola.

Problem 2: \( y > -x^2 + 4x - 3 \)

Step 1: Rewrite the quadratic (or find roots)

First, factor or find vertex. Let's find roots by setting \( -x^2 + 4x - 3 = 0 \) → \( x^2 - 4x + 3 = 0 \) → \( (x - 1)(x - 3) = 0 \), so roots at \( x = 1 \) and \( x = 3 \).
The vertex form: \( -x^2 + 4x - 3 = -(x^2 - 4x + 3) = -(x^2 - 4x + 4 - 1) = -(x - 2)^2 + 1 \).
Since \( a = -1 < 0 \), the parabola opens downward, vertex at \( (2, 1) \).

Step 2: Determine the boundary line

Inequality is \( y > -(x - 2)^2 + 1 \), so boundary is \( y = -(x - 2)^2 + 1 \) (dashed line, because \( > \) does not include equality).

Step 3: Test a point to shade the region

Test the origin \( (0, 0) \):
\( 0 > -(0 - 2)^2 + 1 \)?
\( 0 > -4 + 1 \)?
\( 0 > -3 \), which is true. So shade the region above (but not including) the downward-opening parabola.

Problem 3: \( y \leq -3(x + 1)^2 + 10 \)

Step 1: Analyze the parabola

The quadratic is in vertex form: \( y = -3(x + 1)^2 + 10 \).
\( a = -3 < 0 \), so it opens downward, vertex at \( (-1, 10) \).

Step 2: Determine the boundary line

Inequality is \( y \leq -3(x + 1)^2 + 10 \), so boundary is \( y = -3(x + 1)^2 + 10 \) (solid line, because \( \leq \) includes equality).

Step 3: Test a point to shade the region

Test the origin \( (0, 0) \):
\( 0 \leq -3(0 + 1)^2 + 10 \)?
\( 0 \leq -3 + 10 \)?
\( 0 \leq 7 \), which is true. So shade the region below (including) the downward-opening parabola.

Final Graphing Summary:

1. \( y \geq x^2 + 4x - 5 \): Solid upward parabola, shade above.
2. \( y > -x^2 + 4x - 3 \): Dashed downward parabola, shade above.
3. \( y \leq -3(x + 1)^2 + 10 \): Solid downward parabola, shade below.

(Note: For actual graphing, plot the vertex, roots (if needed), draw the boundary, and shade the correct region.)

Answer:

Step 1: Analyze the parabola

The quadratic is in vertex form: \( y = -3(x + 1)^2 + 10 \).
\( a = -3 < 0 \), so it opens downward, vertex at \( (-1, 10) \).

Step 2: Determine the boundary line

Inequality is \( y \leq -3(x + 1)^2 + 10 \), so boundary is \( y = -3(x + 1)^2 + 10 \) (solid line, because \( \leq \) includes equality).

Step 3: Test a point to shade the region

Test the origin \( (0, 0) \):
\( 0 \leq -3(0 + 1)^2 + 10 \)?
\( 0 \leq -3 + 10 \)?
\( 0 \leq 7 \), which is true. So shade the region below (including) the downward-opening parabola.

Final Graphing Summary:

1. \( y \geq x^2 + 4x - 5 \): Solid upward parabola, shade above.
2. \( y > -x^2 + 4x - 3 \): Dashed downward parabola, shade above.
3. \( y \leq -3(x + 1)^2 + 10 \): Solid downward parabola, shade below.

(Note: For actual graphing, plot the vertex, roots (if needed), draw the boundary, and shade the correct region.)