Chapter 1, Problem 5TP

Solution

To choose: the correct inequality to represent the given graph.

The correct inequality to represent the given graph is B.

Given information: 

The given inequality are

A $y\ge -x^2-x+6$_.

B $y\ge x^2+x-6$

C $y>2x^2+2x-12$

_D.$y\ge -2x^2-2x+12$

The given graph of the quadratic inequality is

  

Formula used:

Standard equation of a parabola is of the form $y=a{x}^2+bx+c$

If a quadratic inequality is less than or greater than type then the outline of the parabola is a dashed line.

If a quadratic inequality is less than equal to or greater than equal to type then the outline of the parabola is a solid line.

If $a>0$ then parabola is open upwards

If $a<0$ then parabola is open downwards.

Calculation:

The outline of the parabola is a solid line.

So, the inequality is greater than equal to type.

The vertex is at $\left(-0.5,-6\right)$ .

Now, check for $x$ -coordinate of the vertex.

  $-\frac{b}{2a}=-0.5$

Now, check for $y$ -coordinate of the vertex.

  $y=-6$

Here, $a=1>0$

The parabola is open upwards

Now, take $\left(0,0\right)$ as check point.

  $\begin{array}{l}0\ge 0-6\\ 0\ge -6\end{array}$

Which is false.

So, the parabola is satisfied by $\left(0,0\right)$ .

The shaded region will be inside and the outline of the parabola.

The parabola is open upwards as $a=1>0$

The correct inequality to represent the given graph is $y\ge x^2+x-6$_.