Chapter 9.2, Problem 42HP

a.

To determine

TO-DETERMINE whether the given function is sometimes, always or never true .

Answer to Problem 42HP

The given statement is never true for value of $k\ne 0$ .

Explanation of Solution

Given information :

The function $y=x^2+k$ has its vertex at the origin.

A quadratic equation would always form a graph that is a parabola. In this context, the vertex of the parabola would be in the origin only if the function is in the form of $y=a{x}^2$ or $y=-a{x}^2$ .

Here ‘a’ is the coefficient of $x^2$ and can hold any value, positive or negative. If there any other variables or constants added to these equations, the vertex would shift to some other point on the graph.

b.

To determine

TO-DETERMINE whether the given function is sometimes, always or never true .

Answer to Problem 42HP

The given statement is always true.

Explanation of Solution

Given information :

The reflection of the graph of $y=a{x}^2$ over the x-axis would have the same width as $y=a{x}^2$ .

The reflection of a function over the x-axis can be derived by multiplying the original function by -1. This means that the signs before the various variables and constants would be reversed. By doing this, there will be a reflection of the parent graph over the x-axis.

However, since the numeric value of the coefficients do not change, the width would remain unaltered.

c.

To determine

TO-DETERMINE whether the given function is sometimes, always or never true .

Answer to Problem 42HP

The given statement is never true.

Explanation of Solution

Given information :

The graph of the equation of $y=x^2+k$ where $k\ge 0$ and the graph of the equation with vertex at $\left(0,-3\right)$ have the same maximum or minimum point.

The graph of the quadratic $y=x^2+k$ where $k\ge 0$ would always have a minimum. This is because $x^2$ has a positive sign. Also, since $k\ge 0$ , the minimum would either lie on the origin or on any point on the y-axis, towards the positive side.

Had $k\le 0$ or $k=-3$ , there could have been a possibility that maximum or minim could lie at $\left(0,-3\right)$ . However, in this case, the given statement is never true.