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Chapter 11: Problem 13

\(3-16=\) Graph the inequality. $$ y>x^{2}+1 $$

Short Answer

Expert verified
Graph the parabola \(y = x^2 + 1\) as a dashed line and shade above it.

Step by step solution

01

Understand the Inequality

The inequality we are dealing with is \( y > x^2 + 1 \). This is a quadratic inequality where the right side is a quadratic expression \( x^2 + 1 \). Our goal is to graph the region where the value of \( y \) is greater than the value of this quadratic expression.
02

Graph the Parabola

Graph the parabola \( y = x^2 + 1 \). This serves as the boundary for our inequality. The graph of \( y = x^2 + 1 \) is a standard upward-opening parabola that is shifted one unit upwards along the y-axis.
03

Determine the Boundary Type

Since the inequality is \( y > x^2 + 1 \) (and not \( y \geq x^2 + 1 \)), the boundary line (parabola) \( y = x^2 + 1 \) is dashed. This indicates that points on the parabola are not included in the solution set.
04

Shade the Correct Region

Shade the region above the parabola \( y = x^2 + 1 \). This is where \( y \) is greater than \( x^2 + 1 \). You can test a point above the parabola (like (0, 2)) to confirm that the inequality holds, ensuring you're shading the correct area.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Inequalities
A quadratic inequality involves an algebraic expression with a quadratic term, which is an expression that includes a squared variable, such as \( x^2 \). In our example, the inequality given is \( y > x^2 + 1 \). Here, the goal is to find the solution set of \( y \) values that are greater than \( x^2 + 1 \). Understanding this is crucial, as it's not just about solving for x or y -- it鈥檚 about determining the range of solutions that satisfy the inequality within a graphing context.
  • A quadratic inequality can take forms like \( y < x^2 + bx + c \) or \( y > x^2 + bx + c \).
  • These inequalities define regions on the coordinate plane to graphically solve them.
  • The solution is often a region, not a specific set of values.
In essence, solving a quadratic inequality graphically involves drawing and analyzing curves called parabolas that represent the boundary of the solution set.
Parabola Graphing
Graphing a parabola is an essential step when dealing with quadratic inequalities. A parabola is a symmetrical curve represented by a quadratic equation such as \( y = x^2 + 1 \). To graph it, you follow these steps:
  • Identify the vertex: For \( y = x^2 + 1 \), the vertex is \( (0, 1) \), since it鈥檚 given by the lowest point on the curve when it opens upward.
  • Determine the direction: An upward opening happens because the coefficient of \( x^2 \) is positive.
  • Draw the axis of symmetry: A vertical line through the vertex at \( x = 0 \).
  • Sketch the parabola: Start at the vertex (0,1) and plot additional points to form a curve.
For inequalities like \( y > x^2 + 1 \), the boundary is represented by a dashed line to indicate that the values on this line do not satisfy the inequality (since \( y \) must be greater, not equal).
Inequality Region Shading
Shading the correct region is vital for showing the solutions to a quadratic inequality on a graph. In the case of \( y > x^2 + 1 \), you need to shade the region above the parabola because any point in this area satisfies the inequality.
  • The inequality sign \( > \) directs you to shade above the parabola.
  • Use a dashed boundary to indicate that points on the line are not part of the solution set.
  • Pick a test point such as (0, 2). Substitute into the inequality: \( 2 > 0^2 + 1 \), which simplifies to \( 2 > 1 \). This holds true, confirming the correct region is above the parabola.
Shading communicates which parts of the graph satisfy the inequality. It鈥檚 an essential skill because it visually differentiates between points that satisfy and those which do not, using logic from inequalities.

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Most popular questions from this chapter

Solve the system of linear equations. $$ \left\\{\begin{aligned} y-z+2 w =0 \\ 3 x+2 y +w=0 \\ 2 x +4 w=12 \\\\-2 x &-2 z+5 w= 6 \end{aligned}\right. $$

Show that $$ \left|\begin{array}{lll}{1} & {x} & {x^{2}} \\ {1} & {y} & {y^{2}} \\ {1} & {z} & {z^{2}}\end{array}\right|=(x-y)(y-z)(z-x) $$

Use Cramer鈥檚 Rule to solve the system. $$ \left\\{\begin{aligned} 5 x-3 y+z =6 \\ 4 y-6 z =22 \\ 7 x+10 y \quad=-13 \end{aligned}\right. $$

Find the determinant of the matrix, if it exists. $$ \left[\begin{array}{rr}{0} & {-1} \\ {2} & {0}\end{array}\right] $$

\(47-50\) . Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, rounded to one decimal place. $$ \left\\{\begin{array}{l}{y \geq x^{3}} \\ {2 x+y \geq 0} \\ {y \leq 2 x+6}\end{array}\right. $$
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