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]]>We will assume you are given the vertex form of the parabola. Say you have y = 3(x-1)² + 2. We are asked to find the focus. How do we go about it?

The formula we need in this case is F(h, k+p). Recall that the vertex is V(h,k) when y = a(x-h)²+k. Because we have the equation in vertex form already, we know the vertex is at V(1,2). The only think we need now is p, where p = 1/(4a). As a = 3, in this case, p = 1/(4*3) = 1/12. So F(h,k+p) = (1, 2+1/12) = (1, 25/12). And so, the focus is the point F(1, 25/12).

Thus, to find the focus of the parabola, we (1) find the vertex, V(h,k); then (2) compute p = 1/(4a); then (3) compute F(h,k+p).

Here’s another example.

**Example 1:** Find the focus of the parabola y = -2(x+4)²-1.

**Solution:** Because y = -2(x+4)²-1, the vertex is the point V(-4,-1). Next we calculate p = 1/(4a), noting that a = -2, in this case. When we do we find p = -1/8. Finally the focus is the point (-4, -1 + -1/8) = (-4, -9/8). ■

Sometimes we are given the equation of the parabola in standard form and must either convert the equation to vertex form or use formulas to compute the vertex.

**Example 2:** Find the focus of the parabola y = (1/2)x² – 5.

**Solution:** Note that, in this example b = 0, and that any time that b = 0, the standard form and vertex form of the equation are identical. But in this case, we will compute the vertex using the formulas we would use if the vertex form of the equation were not also given to us.

The vertex formula gives us h = -b/(2a) and k = (4ac-b²)/(4a). In this case, a = 1/2, b = 0, and c = -5. Thus h = 0 and k = -5 and the vertex is the point (0, -5). With a = 1/2, then p = 1/2, so that F is the point (0, -5+1/2) = (0, -9/2). ■

For more advanced students who wish to prove the formula for the focus, here is an outline of one way to do it.

Next, because we know y is quadratic in x, write the coefficients of this quadratic as A, B, and C, and set out to write m, n, and t in terms of A, B and C. We encounter a system of three equations in three unknowns. Once we solve each m, n and t, in terms of A, B, and C, we obtain the usual relationships by defining a point V(h,k), called the vertex, where h = -B/(2A), k = (4AC-B²)/(4A). Then if p = 1/(4A), then focus, F(m,n) is given by m = h; n = k+p, and directrix is given by y = k-p.

When we’re given the vertex and a point on the parabola, and asked to find the focus, we’re going to want to use a the following process. First, we write the equation of the parabola in vertex form, to the extent we can. We will be able to write the entire vertex form of the equation except for our leading coefficient, a. Next, we’ll use the point that we’re given as a solution to this equation. This means we can substitute the point (x,y) in for the x and y values in the vertex equation. Then we solve for a. After we have a, we can compute p = 1/(4a), and F(h,k+p), which will give us our focus.

Here’s an example of how we go about it.

**Example:** Given the vertex is at V(1,2) and the point (3,4) is on the parabola, find the coordinates of the focus.

**Solution:** First use the vertex to write the equation of the parabola in vertex form to the fullest extent we can. Recall vertex form of the parabola is y = a(x-h)²+k. Because we are given the vertex is V(1,2), the vertex form of the equation of our parabola is y = a(x-1)² + 2.

Next, we use the fact that the point (3,4) is a solution to this equation to find the value of a. We do this by plugging the values x=3 and y=4 into our equation: 4 = a(3-1)²+2. Then we solve to find that a=1/2.

Because we now have the value of a, we can easily find p=1/(4a)=1/2. So, F(h,k+p) = F(1,2+1/2)=F(1,5/2), and our focus is the point F(1,5/2). ■

**Did we answer your question about finding the focus of a parabola? Share your thoughts in the comments.**

There are a couple of ways we can determine the direction the parabola opens. The a-value is one way. If a > 0, then the parabola opens upward. If a < 0, then the parabola opens downward. Recall that when the parabola is given in standard form, y = ax² + bx + c, the a-value is the leading coefficient.

**Solution:** In this example, a = 3, and because 3 > 0, the parabola opens upward. ■

The parabola can also be given in vertex form. Recall the vertex form of a parabola is given as y = a(x-h)² + k. We still use the a-value and use the same test when the equation of the parabola is given to us in vertex form.

**Solution:** In this case, a = – 2 < 0. So this parabola opens downward. ■

Another way we can determine a parabola’s orientation is by knowing the *relative vertical position* between the focus and directrix. When the focus is above the directrix, then the parabola opens upward. When the focus is below the directrix, the parabola opens downward.

**Solution:** Because the directrix is y = 7 and the y-coordinate of the focus is y = 4, the directrix is above the focus. Thus, the parabola opens downward. ■

For you fancy-pants calculus students, you do not need to know anything about parabolas in general to know what direction it opens. Given the equation of the parabola, in either standard form or vertex form, you can perform a test for concavity by taking the second derivative of the given function. On a quadratic polynomial, the second derivative always yields a constant. When it yields a positive constant, we know the original function is everywhere concave up. When it yields a negative constant, the original function is everywhere concave down. In other words, when the second derivative of the equation of our parabola is positive, the parabola opens upward; when it is negative the parabola opens downward. That’s all there is to it. Here’s how it works in practice.

**Solution:** Because y = 2x² + 4x – 1, then y’ = 4x + 4, and y”= 4. Because y” = 4 > 0, the parabola is everywhere concave up. Thus, the parabola opens upward. ■

Here is an example of determining the parabola’s orientation with the concavity test when the original equation is given in vertex form.

**Solution:** If y = -2(x-1)² + 3, then y’ = -4(x-1), and y” = -4. Because y” = -4 < 0, the parabola is everywhere concave down. So the parabola opens downward. ■

Here’s what we learn when we apply the concavity test to the general quadratic polynomial.

**Solution:** Because y = ax² + bx + c, y’ = 2ax + b, and y”= 2a. Note that y” > 0 if a > 0, and y” < 0 if a < 0. So the value of a determines whether the parabola is everywhere concave up or everywhere concave down. Thus, the parabola opens upward whenever a > 0. The parabola opens downward whenever a < 0. ■

**Did we answer your question about parabola orientation? Share your thoughts in the comments.**

Any time we are given only one or more of the following:

- Focus
- Vertex
- Axis of Symmetry
- Directrix

and are asked to find one or more of the others, we must first determine the orientation of the parabola, if we are not already told what it is (as we usually are not).

In the cases when we are given either the axis of symmetry or directrix, we will be able to tell right away the parabola orientation because vertically oriented parabolas have vertical axes of symmetry and horizontal directrices. Therefore we will be given equations for a vertical line (“x=[a constant]”) in the case of the axis of symmetry, and an equation of a horizontal line (“y=[a constant]”) in the case of the directrix. When the parabola is horizontally oriented, we will have been given horizontal axes of symmetry (“y=[a constant]”) and/or vertical directrices (“x=[a constant]”).

In the case that we are given either only the focus or only the vertex of a parabola, we will not be able to determine the parabola’s orientation — we need both, or at least some other piece of information in addition only one of these. When we are given both, we can quickly tell the orientation of the parabola by the repeated value in either the x-coordinates of the focus and vertex or the y-coordinates of the focus and vertex.

If the repeated value occurs in the x-coordinates of the focus and vertex, this means the axis of symmetry is vertical, and the parabola is vertically oriented. (Try to visualize this. **See Figure A, Example 1.**)

If the repeated value occurs in the y-coordinates of the focus and vertex, the axis of symmetry is horizontal, and the parabola is horizontally oriented. (Try to visualize this. **See Figure A, Example 2.**)

So for example, if we are given that a parabola’s focus is F(3,-2) and vertex is V(-3,-2), then the repeated value occurs in the y-coordinates; the axis of symmetry is therefore horizontal, and the parabola is likewise oriented horizontally.

Once we have found the orientation of the parabola, we can find the directrix in a couple of ways. Possibly the most straight forward way is to use the midpoint formula, given that the vertex is midpoint between the focus and a collinear point on the directrix. **See Figure B. ** So for instance, if we have V(3,1) and F(3,4), we find d in the point D(3,d):

(d+4)/2 = 1; d+4=2; and d=-2.

Thus, D(3,-2) is a point lying on the directrix. Because we determined the parabola was vertically oriented, the directrix is the horizontal line passing through D(3,-2), which is y=-2. Here is another example:

**Example:** If the vertex of a parabola is V(3,1) and its focus is F(1,1), find the directrix of the parabola.

**Solution:** We are asked to find the directrix given the vertex and focus of the parabola, so first we determine the orientation of the parabola — which we find through identifying the repeated component value present in the vertex and focus; this being y=1, the y-coordinate, we know the axis of symmetry is horizontal, the parabola is horizontally oriented, and the directrix is vertical.

Next, we find the directrix by using the fact that the vertex V(3,1) is the midpoint between the focus F(1,1) and a collinear point on the directrix, D(d,1) Thus:

3=(1+d)/2; 6=1+d; and d=5.

Because we determined the directrix is vertical, the point D(5,1) on the directrix gives us the equation of the directrix x=5.

**Have something to add to this walkthrough? Share it in the comments.**

**Additional Problems**

(1) Explain how we know that F(-1,2) and V(-1,4) are from a vertically oriented parabola.

(2) How do we know that the parabola with axis of symmetry y=-3/4 is horizontally oriented?

(3) Why do we know that the directrix x=9 is from a horizontally oriented parabola?

(4) Show that if the focus of a parabola is at F(2,3) and its directrix is at y=-3, the vertex is at V(2,0).

(5) Show that if F(2,1/2) and V(3,1/2) the directrix is given by x=4.

(6) Show that if F(-5,1) and V(-5,10), the directrix is given by y=19.

]]>Watch two things in these types of problems. First, the geometry of the situation — i.e. the orientation of the parabola and its various parts; and second, make sure your formulas are properly applied, as there are many k’s, p’s and a’s floating around and are a lot to contend with.

**SEE ALSO: How to find the directrix of a parabola, given its vertex and focus.**

The geometry in this case is straight forward — we have a parabola that is opening upward in the plane. This is because we are quadratic in x (as opposed to y — see what to do here when it’s not), and the leading coefficient a is positive. **See Figure A.**

Because the parabola is opening upward, our focus will have the highest y-coordinate, the vertex will be south of the focus, and finally the directrix will be furthest south of them all.

With that in mind, to actually find the vertex, focus, and directrix, you can either transform your y=x²-6x+15 into vertex form, or you can leave it, and use formulas for these from standard form. The consideration that I make when deciding which to use is whether the standard form is “easily” made into vertex form. When I look at y=x²-6x+15, I do not see a quick way to turn it into vertex form through factoring (meaning we would have to complete the square). Thus, I would keep y=x²-6x+15 in standard form and use formulas from there. To help us with our calculations, we’ve put together a quick reference “cheat sheet” in **Figure B.**

We’re also solving this problem in the context of multiple choice, which is great because the answer is somewhere already on the paper! It’s even better because sometimes you won’t have to derive the entire answer to rule out certain ones. (See the **Test Taking Tips** below, for some tips about how to use multiple choice answers to your advantage.)

First, we’ll work on the vertex. The vertex x-coordinate is given by x=h=-b/(2a). Recall that when in standard form, the convention is to use coefficients a, b, and c as in ax²+bx+c. Thus, a=1, b=-6, and c=15 in our example, and the x-coordinate of the focus is:

x = -b / (2a) = – (-6) / (2*1) = 6/2 = 3.

To find the y-coordinate of the vertex, k, we can either use the formula (4ac-b²)/(4a) in **Figure B** or plug the x-coordinate we found into y=ax²+bx+c. We’ll do the latter, using x=3 we just found:

y = ax² + bx + c = 1*(3)² – 6 (3) + 15 = 9 – 18 + 15 = 6.

Thus, the vertex is (3,6).

**TEST TAKING TIP #1:** Notice here that we have only one answer that has vertex (3,6), and that answer is A. Thus, at this point, I would double check my calculations very carefully (as always, but especially in this case, as we would be betting the house on it), and pick that answer. Then move on to the next question from there. I’ll do this especially when the test is timed. We’ll continue the rest of the calculations, however, to be sure you know how to do the rest.

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The focus lies on the axis of symmetry of the parabola, and has the y-coordinate k+1/(4a). Because we just found the vertex to be (3,6), we know the axis of symmetry to be x=3, and the focus lies on that line. See **Figure B.** Its x-coordinate is therefore x=3. Because in our problem a = 1, the y-coordinate of the focus is 1/(4a) = 1/(4*1) = 1/4 north of the vertex. Thus, y = 6 + 1/4 = 6.25, and our focus is (3,6.25), which is what we were expecting from our selection of multiple choice answer A.

**TEST TAKING TIP #2:** Note that we know that the vertex x-coordinate and focus x-coordinate are always the same. So even before we made any calculations in this problem, once we knew the parabola opened upward, we could rule out multiple choice answer D, as that option presents a vertex and focus with different x-coordinates.

Finally, we can find the directrix of a parabola by noting that it will be a horizontal line and south of the vertex of the upward opening parabola, as we said above. Once again, see **Figure B.** Once you know the y=coordinate of the vertex, k, it is given by y = k – p, where p = 1/(4a). Thus, as we calculated for the focus, above:

p = 1/(4a) = 1/(4*1) = 1/4

and the directrix is y = 6 – 1/4 = 5.75, once again confirming our answer of multiple choice option A.

**Have something to add to this walkthrough? Share it in the comments.**

In this problem, we walk through a word problem involving an accounting application in which we must calculate changes in owner’s equity based on information given about its assets and liabilities and their changes over the year.

These problems can be tricky. The thing to keep in mind is that balance sheets reflect the financial state of a firm at a single point in time. Not a period of time, like income statements or cashflow statements. When you’re given information about change in assets, for example, it is not listed on the balance sheet, itself; you have to find another way to diagram it. In the tutorial, I used arrows to indicate the change, but anything will do, so long as it doesn’t end up on the sheet itself.

Would you like to see additional tutorials on application problems in accounting? What would you like to see?

]]>This is a video session on how to multiply and divide fractions, as how students generally encounter it in Pre-Algebra and early Algebra courses.

Notice that the illustration starts off with some relatively simple multiplication problems, and ultimately ends with some more complex division of mixed numbers. This wasn’t an accident. It illustrates that these types of problems can be as easy as a 3 step process:

- Multiply across to calculate the new numerator and denominator;
- Use the prime factorization process to identify and eliminate common factors in the numerator and denominator; and
- Multiply-out to find the final answer;

or as complex as a 6 step process:

- Convert mixed numbers to improper fractions;
- Convert “division by a fraction” to “multiplication by the reciprocal of the fraction”;
- Multiply across to calculate the new numerator and denominator;
- Use the prime factorization process to identify and eliminate common factors in the numerator and denominator;
- Multiply-out to find the final answer, possibly as an improper fraction; and
- Convert the improper fraction to a mixed number, if necessary.

When we teach this skill, we use a building approach in which we make sure that you feel comfortable first with problems requiring the simple process, and then move on to the more complex. From there, we help you become fluent in the simple ones before moving on to master the more complex.

]]>In this problem, we walk through a question on calculating compound interest. In the explanation video, we spend some time on on the compound interest formula, itself, and distinguish it from simple interest scenarios.

**Comparison with Simple Interest: **Notice that at the end of the investment period, you can see the impact of compounding. For example, in Part (a) of this problem, simple interest would yield $1,060.00 whereas we calculated that compounding yields $1,061.36 — or a difference of $1.36 due to the compounding. While this isn’t very much difference in just one year, the difference grows when the compounding is allowed time to grow. In part (b), simple interest in two years yields $1,120.00 compared to $1,126.49 (a difference of $6.49). Part (c), in 5 years, simple interest yields $1,300.00 versus $1,346.86 (a difference of $46.86). And in Part (d), a simple interest investment would yield $1,600.00 after 10 years compared to $1,814.02 with compounding (a difference of $214.02) — significantly more growth than in the simple interest scenario. In fact, you would need a simple interest rate of 8.14% to match the compound growth at 6% in 10 years!

As we mention in our response to this student, we have been experimenting with video responses to questions, and provided one to help see the geometry.

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