Question

(a) Find the equation of the graph which is obtained from the graph of \(y = {e^x}\)and is reflected about the line \(y = 4\).

(b) Find the equation of the graph which is obtained from the graph of \(y = {e^x}\)and is reflected about the line \(x = 2\).

Short answer

1. The equation of the resultant graph is, \(y = {e^{ - x}} + 8\).
2. The equation of the resultant graph is \(y = {e^{ - x + 4}}\).

Step-by-step solution

Given data

The equation is\(y = {e^x}\).

Concept of Vertical and horizontal shifts

Vertical and horizontal shifts:

When\(y = f(x) + c\)and\(c > 0\), shift the graph of\(y = f(x)\)a distance\(c\)units upwards.

When\(y = f(x) - c\)and\(c > 0\), shift the graph of\(y = f(x)\)a distance\(c\)units downwards.

When\(y = f(x) - c\)and\(c > 0\), shift the graph of\(y = f(x)\)a distance\(c\)units towards the right.

When\(y = f(x) + c\)and\(c > 0\), shift the graph of\(y = f(x)\)a distance\(c\)units towards the left.

Obtain the equation from the graph of \(y = {e^x}\)and is reflected about the line \(y = 4\)

(a)

Start with the equation \(y = {e^x}\).

The reflection of the graph of \(y = {e^x}\) about the \(x\)-axis is \(y = {e^{ - x}}\).

The graph \(y = {e^x}\) has \(y\)-intercept at \(y = 1\).

So, its reflection will pass through \(y = - 1\).

To reflect the graph about the line \(y = 4\) add \(8\) to both sides so, the reflected graph will pass through \(y = 4\).

Thus, the equation of the resultant graph is \(y = - {e^x} + 8\).

Obtain the equation from the graph of \(y = {e^x}\)and is reflected about the line \(x = 2\)

(b)

Start with the equation \(y = {e^x}\).

The reflection of the graph of \(y = {e^x}\) about \(y\)-axis is \(y = {e^{ - x}}\).

When the function \(y = {e^{ - x}}\) is drawn then the vertical asymptote is zero.

In order to set \(x = 2\) in the centre of the graph of these two functions, add 4 units from the origin of \(x\)-axis in the right direction.

Thus, the equation of the resultant graph is \(y = {e^{ - x + 4}}\).