Lesson 10 Comprehension Check number 2 Left click your mouse or press a key to advance the slide show.

Graph

In order to graph this equation in two variables we first need to recognize which type of graph this equation represents. Since y is degree one and we have an x that is degree two we have an up-down type of parabola.

To graph parabolas we want to get their equations into vertex form, so that we can identify the location of the vertex. I will do this by completing the square. I will demonstrate by using one of the styles shown in the lesson.

Isolate the x-squared term and the x term, so add 5 to both sides first.

Next, we need to make the coefficient of the x-squared term a one, or else the completing of the square will not work, so we divide both sides by the coefficient of the x-squared term, which in this instance is three.

Now we can get started with completing the square. We take half of the coefficient of x and square it. In this instance the coefficient of x is two. Half of two is one. Square one and we get one. This is the quantity we add to the right hand side to make a perfect square trinomial. Of course to keep balance we must add it to the left side as well.

Next, we factor the perfect square trinomial on the right hand side so that we have the square of a binomial.

Now that we have finished completing the square we can undo what we did earlier, so that we can get y by itself again on one side. Start by subtracting one from both sides.

Next, multiply both sides by three to clear out the denominator.

Finally, subtract five from both sides.

This is the vertex form that most authors use. In my lesson I would instead add three to both sides to get

Slightly different look, but yields the same information. You just read it a little differently from what most authors do. Given either form you should be able to tell that the vertex is the point (-1,-8). With my form you take the opposite of what you see for both h and k (the x and y coordinates of the vertex). In the authors’ form you take the opposite for h, but read the k as it is shown.

Now that we know the vertex, let’s find some other points to help make the graph look better. Start with the y-intercept. To find this we let x = 0, and solve for y. I like to use the standard form to find this, but the vertex form works as well.

Now, we know that the graph crosses the y-axis at -5. As a point the y-intercept is (0,-5).

Next, let’s find the x-intercepts if they exist. Since the vertex is below the x-axis and the parabola opens up we know that the graph will cross the x-axis, so the x- intercepts are real and worth finding. I like to use the vertex form to find these, but the standard form works too.

To find the x-intercepts let y = 0.

To solve for x, first add 8 to both sides.

Then divide both sides by three.

Now use the square root property to undo the squaring of the x-expression.

Notice that the square root is not in simplified form. We do not want fractions inside the radical sign, so we first try using the quotient rule of radicals to split the fraction.

This eliminates one problem, but creates another. If the denominator had a perfect square we could then evaluate the square root and be done, but in this instance we end up with a square root of three in the denominator, which still isn’t simplified, so now we try rationalizing the denominator.

Now the denominator looks good, but we still have one more simplification problem. The radicand in the numerator, 24, has a perfect square as a factor, namely 4, so we need to remove that to finish the simplification process.

Last, but not least we solve for x by subtracting one from both sides.

It is difficult to tell where to place the x- intercepts when we have a radical, so let’s use a calculator to approximate the x- intercepts. I get

Now we have enough points to make a pretty nice looking sketch. Place the vertex and the intercepts on a grid, and connect the dots with a parabola. Hopefully your graph looks like …