Chapter 9.1, Problem 71HP

To determine

To formulate a quadratic equation for the described graph.

Answer to Problem 71HP

The quadratic equation representing the parabola shown in the graph is $y=-x^2+6x+16$

Explanation of Solution

Given information :

Quadratic equation with vertex at $(2,0)$ and a point on the graph is $(5,9)$

Formula used :

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.

The axis of symmetry bisects the parabola into two equal parts. Hence each point on the parabola would have an equal point on the other side of the axis of symmetry

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’ is lesser than 0 hence the graph will have a maximum and will open downwards.

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

In the given case the vertex is at $(2,0)$ and the given point $(5,9)$ on the graph is higher than the vertex which signifies that the parabola is upward facing and the vertex is at the minimum

Formula to compute equation of the axis of symmetry $x=-\frac{b}{2\times a}$ .

Axis of symmetry is

  $x=2$

Calculation :

Equating the two equations above.

  $-\frac{b}{2\times a}=2$

Simplifying the expression.

  $b=-4a$

In the above expression $a$ has to be positive as the parabola is upward facing

Putting $a=1$ .

  $b=-4\times 1$

  $a{x}^2+bx+c$

In quadratic form,

  $y=x^2-4x+c$

Here, c is not known

Putting $x=5$ and $y=9$ in the above expression, corresponding to the point $(5,9)$ in the graph.

  $9=5^2-4\times 5+c$

Simplifying.

  $c=4$

Therefore, the quadratic equation is $y=x^2-4x+4$