Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
3. Section 8.3
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Exercise 120 Page 518

Practice makes perfect
a In order to find the value of f(1), let's draw the line x= 1 and check where it intersects the graph of our function.

Now that we found where the line x=1 intersects our function, let's find the y-coordinate of the point of intersection by moving horizontally until we hit the y-axis.

We reached the y-axis at y = -3, which means that -3 is the y-coordinate of the point of intersection of the graph of our function with the line x = 1. In other words, f( 1) = -3.

b In order to find the value of f(0), let's draw the line x= 0 and check where it intersects the graph of our function.

Notice that the y-coordinate at this point is the y-intercept of the function. Therefore, f( 0) = -4.

c In order to find the value or values of x such that f(x) = 4, let's draw the line y= 4 and check where it intersects the graph of our function.

Now that we can see where the line y=4 intersects our function, let's find the x-coordinates of these points by moving vertically from the points of intersection until we hit the x-axis.

This time we cannot determine the exact x-coordinates of our points, so we have to approximate them. It seems that we hit the x-axis at x ≈ -2.8 and x ≈ 2.8. Therefore, we can satisfy f(x) = 4 when x ≈ -2.8 and x ≈ 2.8.

d In order to find the number or numbers x such that f(x) = 0, let's draw the line y= 0 and check where it intersects the graph of our function.

Notice that this line is the x-axis. In other words, the x-coordinates of these points are the x-intercepts of the function. Let's identify them.

When f(x) = 0, x = -2 and x = 2.