Chapter 9, Problem 2PFA

To determine

To evaluate the value of b when equation is written in standard form.

Answer to Problem 2PFA

  $b=10$

Explanation of Solution

Given information :

Original quadratic equation = $f(x)=x^2$

Reflected across x -axis, translated 5 units to the right.

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’ 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.

The graph represented by $g(x)={(x-h)}^2$ is equivalent to the graph of $f(x)=x^2$ that has been translated horizontally. The direction of translation or movement of the graph is dependent on the value of $h$ .

If $h>0$ , the graph of $f(x)=x^2$ is translated $\left|h\right|$ units to the right.

If $h<0$ , the graph of $f(x)=x^2$ is translated $\left|h\right|$ units to the left.

Therefore, the original function $f(x)=x^2$ is transformed to $f(x)=-{(x-5)}^2$ ;

Simplifying the expression;

  $f(x)=-(x^2-2\times x\times 5+25)$

Therefore, the function in standard form $y=a{x}^2+bx+c$ is.

  $f(x)=-x^2+10x-25$

The value of b is 10