Chapter 9, Problem 4CR

To determine

To select options that describe the graphed function.

Answer to Problem 4CR

Option A & E .

Explanation of Solution

Given information :

Option A. The range is $\left\{y\right|y\ge -4\}$

Option B. The y- intercept is 3.

Option C. The maximum value is -4

Option D. The equation of the axis of symmetry is $x=-4$

Option E. The x- intercepts are -1 and 3.

:

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

Highest point (if $a<0$ ) and be called the maximum. Here, the graph opens downwards.

Or, lowest point (if $a>0$ ) and can be called the minimum. Here the graph opens upwards.

In this case, the graph opens upwards hence the function has a minimum value

A parabola always points to infinity, either negative or positive.

Option A. The range is $\left\{y\right|y\ge -4\}$

This option is correct as minimum value as shown in graph is -4.

Option B. The y- intercept is 3.

The option is incorrect as the y- intercept is approximately at -3 and not 3.

Option C. The maximum value is -4

-4 is the minimum value and not the maximum value

Option D. The equation of the axis of symmetry is $x=-4$

The axis of symmetry is to the right of the y- axis and hence cannot be $x=-4$ .

Option E. The x- intercepts are -1 and 3.

As shown in the graph the x-intercepts are -1 and 3, hence this option is correct.