Chapter 9, Problem 16MCQ

To determine

To select from options the equation that correctly describes the graph.

Answer to Problem 16MCQ

Option D. $y=-2x^2+1$

Explanation of Solution

Given information :

Option A. $y=-2x^2$

Option B. $y=x^2-1$

Option C. $y=2x^2+1$

Option D. $y=-2x^2+1$

From the graph shown, it is evident that the vertex is at $(0,1)$ and the graph opens downwards and therefore has a maximum value.

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, ‘a’ has to lesser than 0 hence the graph will have a maximum and will open downwards.

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

Option B. $y=x^2-1$ and Option C. $y=2x^2+1$ are eliminated since a has a positive value which is incorrect.

Option A. $y=-2x^2$

Here, the graph is reflected across the x- axis but not translated as shown in the graph. Hence we reject this option.

Option D. $y=-2x^2+1$

Here, the graph is reflected across the x- axis and translated upwards as shown in the graph. Hence, this equation is the one shown in the graph.