Question

Graph the solution set of each system of inequalities or indicate that the system has no solution.﻿{y≥x2−1x−y≥−1\begin{cases}
y \geq $$x^2 - 1$$ \
x - y \geq -1
\end{cases}
{y≥x2−1x−y≥−1​

Accepted Answer

Step 1: Analyze the first inequality $$y \geq \frac{x$$^{2}$$}{2} + 1. $$This represents the region on or above the parabola $$y = \frac{x$$^{2}$$}{2} + 1. $$The parabola opens upward with vertex at $$(0,1). $$Step 2: Analyze the second inequality $$4x - 2y \geq -2. $$Rearrange it to slope-intercept form by isolating $$y$$: add $$2y to $$both sides and add $$2 to $$both sides to get $$4x + 2 \geq 2y$$, then divide both sides by 2 to get $$2x + 1 \geq y$$, or equivalently $$y \leq 2x + 1. $$This represents the region on or below the line $$y = 2x + 1. $$Step 3: Graph both boundary curves: the parabola $$y = \frac{x$$^{2}$$}{2} + 1$$ and the line $$y = 2x + 1. $$Use solid lines because the inequalities include equality (\geq and \leq). Step 4: Determine the solution set by finding the intersection of the two regions: the area on or above the parabola and on or below the line. This overlapping region satisfies both inequalities simultaneously. Step 5: To better understand the solution set, find the points of intersection between the parabola and the line by setting $$\frac{x$$^{2}$$}{2} + 1 = 2x + 1. $$Solve this equation for $$x to $$find the intersection points, which will help in accurately shading the solution region.

Graph the solution set of each system of inequalities or indicate that the system has no solution.
$\begin{cases}\)y \(\geq\) x^2 - 1 \(\x\) - y \(\geq\) -1\(\end{cases}$

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Key Concepts

Graphing Quadratic Inequalities

Graphing Linear Inequalities

Solution Set of a System of Inequalities

Graph the solution set of each system of inequalities or indicate that the system has no solution.{y≥x2−1x−y≥−1\(\begin{cases}\)y \(\geq\) x^2 - 1 \(\x\) - y \(\geq\) -1\(\end{cases}\){y≥x2−1x−y≥−1