Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Exponential Functions
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Exercise 31 Page 457

Use the calculator to graph two different functions, one for each side of the equation.

About - 2.96 and about 1.34.

Practice makes perfect

To solve the equation by graphing, we first have to create two functions. Each side of the given equation will become a function. x+3=3^x ⇒ lf(x)=x+3 g(x)=3^x Now we want to draw these graphs using a calculator. We can do this by pressing the Y= button and typing the equations in the first two rows. We will use Y_1 for f(x) and Y_2 for g(x).

Fönster med funktioner

Before we graph these functions, notice that in the second one the exponent of 3 is x, so the function will grow very quickly. In order to accommodate this issue, we should change the viewing window. Let's restrict the x-axis. - 5 < x < 5 Also, we can change the scale of the y-axis to increase by 5. We can do this by pushing WINDOW.

Fönster med funktioner
Fönster med funktioner

There are two points of intersection of these lines. To find these points we can use the intersect option, which we get by pushing 2nd and TRACE. Now, we select both graphs and provide the calculator with a guess as to where the intersection might be.

Fönster med funktioner

We will have to do this twice, once for each point of intersection.

Fönster med funktioner
Fönster med funktioner

The solutions to the equation are x ≈ - 2.96 and x ≈ 1.34.