Posts tagged with: parabolas

Huge shortcut here if you just know that for a parabola in standard ax^2 + bx + c form, the x-coordinate of the vertex will be at –b/(2a). In this case, that means it’s at –3/(2(–6)) = 3/12 = ¼.

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PWN the SAT Parabolas drill explanation p. 325 #10: The final way to solve: If we are seeking x=y, since the point is (a,a), why can you set f(x) = 0? You start out with the original equation in vertex form, making y=a and x=a, but halfway through you change to y=0 (while x is still = a). How can we be solving the equation when we no longer have a for both x and y?

For everyone else’s context, here’s the problem:

Now, be careful! I am not changing to y = 0 in that algebraic solution in the back of the book; I am subtracting a from both sides! Note how the a term on the right changes from –22a to –23a.

 

Basically, the question is: how many seconds is h greater than 21? (This tennis ball is being thrown on a planet other than Earth, by the way. I challenge anyone to throw a tennis ball that stays in the air anywhere near as long as this one does.)

To figure it out, solve for the two times the equation equals 21. The first time will be when the ball crosses into view, and the second one will be the time the ball falls out of view. The time in between is the answer you want.

So there you go–the ball is at height 21 at 1 second and at 21 seconds. There are 20 seconds between there, so that’s the amount of time the ball is visible to the kids on the roof.

The other way some folks might choose to solve this is by graphing. If you graph the given function and also a horizontal line at y = 21, you get a nice visual of the ball’s flight, which helps make the 20-second window the ball is visible more intuitive.

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A question from the May 2018 SAT (Section 4 #18)

kx + y = 1
y = -x² + k

In the system of equations above, k is a constant. When the equations are graphed in the xy-plane, the graphs intersect at exactly two points. Which of the following CANNOT be the value of k?

A. 3
B. 2
C. 1
D. 0

Because this comes in the calculator section, you might think that my recommendation is to graph them all. Indeed, if you go that route, you’ll get the answer, and it won’t take that long! However, if you notice that making k equal zero eliminates the x term in the first equation, you might not need to bother graphing anything else.

When k=0, the equations become y=1 and y=-x^2. The former is just a horizontal line above the x-axis, and the latter is a downward-facing parabola that intersects the origin. Those will never intersect, so zero cannot be the value of k.

Hi Mike, Can you please explain Question 11, Test 6, section 3 ? I know the parabola opens downward, but I’m confused after that. Thanks.

You’ve already got the first part—when you add a negative to the front of a parabola you’ll flip it vertically. Since this parabola begins facing up, the negative flips it so it faces down.

From there, I think it’s helpful to think about this in terms of how functions shift. For some function f(x) that’s graphed on the xy-plane:

  • f(x)+1 ⇒ (graph moves UP one)
  • f(x)–1 ⇒ (graph moves DOWN one)
  • f(x+1) ⇒ (graph moves LEFT one)
  • f(x–1) ⇒ (graph moves RIGHT one)

That applies here. You’re basically taking the ax^2 function and applying a few shifts (and a flip). Check it out–if you start by assuming a function p(x)=ax^2, you can build the function in this question by shifting that original function:

    \begin{align*}p(x)&=ax^2\\-p(x)&=-ax^2\\-p(x-b)&=-a(x-b)^2\\-p(x-b)+c&=-a(x-b)^2+c\end{align*}

-p(x-b)+c flips the function p(x) upside-down, then shifts it b units to the right and c units up.

If you hate everything I’ve just said, I have good news! You can also just memorize the vertex form of a parabola (which we basically just derived).

If you have a parabola in the form f(x)=a(x-h)^2+k, then you know it has its vertex at (h, k) and that the sign of a tells you whether the parabola opens up or down. This question basically gives you the vertex form, only it uses b and c instead of h and k. Recognize that and you know right away that the parabola’s vertex is at (b, c).

In PWN p. 159 (p. 157 later printing) #8
In the xy-plane, where a and b are constants, the graphs …

The question does not specify that a and b are positive values. If one or both were negative, wouldn’t that change the answer?

Good question! It turns out, even though my solution assumes a and b are positive, that they don’t need to be. You have two parabolas with vertices on the x = 5 line, one higher than the other. As long as the leading coefficient of the higher one is greater than or equal to the leading coefficient of the lower one, the parabolas will not intersect. C is the answer.

Let’s look at the cases you’re asking about, where one or both of the coefficients is negative. First, say a = 1 and b = –1. In that case, ab, and one variable is negative.

We’re OK there—obviously those parabolas won’t ever intersect.

In the case where both are negative but ab, it turns out we’re STILL fine. Let’s say a = –2 and b = –3. It’s still true that ab, and both coefficients are negative.

Those also won’t ever intersect.

What’s going on here is that as long as ab and both signs are the same, the absolute value of the leading coefficient of the inner parabola will be higher—the inner parabola will be skinnier than the outer parabola!

That’s a lot to think about on one question though—maybe in the next edition I should just make them positive constants!

Test 6 Section 3 #13

I’ll give you two ways to think about this one. Both begin with taking the given equation and putting it in standard form:

    \begin{align*}2x^2-4x&=t\\2x^2-4x-t&=0\end{align*}

The quick and dirty way to go from there (my preferred method because I’m all about graphs) is to recognize that this is a parabola that opens upwards (the first coefficient of positive 2 tells us that) and has a y-intercept of –t. Because a parabola has no real solutions when it doesn’t cross the x-axis, and because only one of the four choices can be correct, the highest y-intercept must be the answer. That will happen when –t is the greatest, which will happen when t is the least. When t = –3, then –t = –(–3) = 3. We’re not going to do better than that, so choice A is the answer.

The mathier way to go is to is to use the discriminant (AKA the part under the square root sign in the quadratic formula). When the discriminant is negative, there are no real solutions.

The discriminant is b^2-4ac. In this case, that’s (-4)^2-4(2)(-t)=16+8t. Of the choices, which value of t makes the discriminant negative? Only choice A, –3, does.

Test 2 Section 4 #7

When you get one of these “equivalent forms” questions, especially in a calculator section, you have lots of options. But before we get to those, let’s just make sure we understand what the question is asking.

“Which of the following equivalent forms of the equation…” means that ALL FOUR CHOICES should be equivalent—if you graph them all you’ll get the same parabola!

“…displays the x-intercepts of the parabola as constants or coefficients?” means that you’re looking for the form that has the actual numbers of both the parabola’s x-intercepts in the equation itself.

Now, here’s where your options come in.

First and fastest, if you know your parabola forms well, you know that the factored form, y=(x-a)(x-b) gives you the x-intercepts (they’re at a and b). Recognize that and you’re down to choice D right away.

If you don’t know that (or if you don’t trust your memory of arcane parabola forms), you can always graph the given equation, see its x-intercepts, and then look for those numbers!

Looks like the x-intercepts are at 2 and 4, so find the choice that has both a 2 and a 4 in it! Only choice D does, so that’s the answer.

PSAT #1, Section 3, #13

General rule: when a question gives you a function and a point on that function, plug the point into the function. In this case, that means plugging (-1, 1) into y=ax^2+bx+c:

    \begin{align*}y&=ax^2+bx+c\\1&=a(-1)^2+b(-1)+c\\1&=a-b+c\end{align*}

And you’re already done! For the parabola to pass through that point, a-b+c must equal 1.

Hi, Mike! Can you explain the second way we can approach question number 6 from the Parabola chapter? (The two points where the higher y-coordinate is also farther from the line of symmetry.) It would be great if you can provide an example. Thank you.

Sure. Here’s the question:

The key, as I say in the solution, is to look for points that are consistent with what you know about the parabola given the equation: that it opens upwards and that its line of symmetry is at x = 6. Because parabolas get wider and wider, points in a parabola that opens up must get higher and higher the farther they get from the line of symmetry.

Therefore, you’re looking for either

  1. Two points with the same y-coordinate that are the same distance from the line of symmetry, or
  2. Two points that are otherwise consistent with the shape of an upwards-opening parabola that’s symmetrical about x = 6.

Choice D does the first one—x = 2 and x = 10 are each 4 away from the line of symmetry at x = 6, and both points have the same y-coordinate of –5.

No choice satisfies the second one (otherwise there’d be more than one possible right answer!) but a choice like (4, 10) and (9, 15) would. (4, 10) is closer to x = 6 and has a lower y-value; (9, 15) is farther from x = 6 and has a higher y-value:

Parabola D in the xy-plane has equation x – 2y^2 – 8y – 11 = 0. Which equation shows the x-intercept(s) of the parabola as constants or coefficients?

A) x = 2y^2 + 8y + 11
B) x = 2(y + 2)^2 + 3
C) x – 3 = 2(y + 2)^2
D) y = – √(x – 3)/2 – 2

The answer is A) which makes me think this is really just a “do you know the definition of this term” kind of question (similar to official practice test 4.4.28). Can you explain for those weak in this? TIA!

When a question asks for an intercept or a vertex or some other notable point “as constants or coefficients,” you’ve got a couple options.

The first is to recognize the form that typically gives that information. For example, if you needed to see the x-intercepts of a vertical parabola (the kind we almost always see), you’d look for answer choices in factored form—e.g., y=(x+2)(x-5)—because that form tells you the x-intercepts. That method works here, but it’s a bit less straightforward than usual because we’re dealing with a horizontal parabola. Basically, the thought process is:

  1. Recognize that the equation gives you a horizontal parabola (it’s got a y^2 term instead of an x^2 term)
  2. Recognize that the properties of a horizontal parabola should be similar to those of a vertical parabola only the x and y terms will be flipped
  3. Know that in a vertical parabola, the standard form—y=ax^2+bx+c—gives you the y-intercept (the c term), so in a horizontal parabola you should be able to get the x-intercept from a similar form
  4. The only answer choice in the x=ay^2+by+c form is A

The second option is to solve for the number being requested and then look for it in the answer choices. In this case, we want x-intercepts. We can always find a graph’s x-intercepts by setting y equal to zero and solving for x.

    \begin{align*}x-2y^2-8y-11&=0\\x-2(0)^2-8(0)-11&=0\\x-11&=0\\x&=11\end{align*}

Now you know you’re looking for an answer choice with an 11 in it. Only choice A has that!

Practice Test 4, Section 3, Number 11 (No Calc)

I love this question because the fastest way to go involves almost no math. You just have to know a little bit about the shapes of lines and parabolas.

First, think about the parabola. Its equation is y=(2x-3)(x+9). From that we know it’s a parabola that opens up (the x^2 term will be positive) and has x-intercepts at \dfrac{3}{2} and -9. You should figure out its y-intercept, too, by plugging in zero for x:

    \begin{align*}y&=(2(0)-3)(0+9)\\y&=(-3)(9)\\y&=-27\end{align*}

Do a very rough drawing of that on your paper (forgive my MS Paint skillz, but your drawing can be sloppier than mine and still be plenty good enough):

Now do a rough drawing of the line. To do that, put it in slope-intercept form:

    \begin{align*}x&=2y+5\\x-5&=2y\\y&=\dfrac{1}{2}x-\dfrac{5}{2}\end{align*}

The important detail there is that the y-intercept is -\dfrac{5}{2}, which is above the parabola’s y-intercept of -27, so you know the line will intersect the parabola twice. Like so:

How do you do Test 5 Section 4 #35?

The area of a rectangular garden will be its length times its width: A=lw. If we know that the length of this garden is 5 feet longer than its width, we can substitute l-5 for w: A=l(l-5). From there, we set the area equal to 104 and solve.

    \begin{align*}104&=l(l-5)\\104&=l^2-5l\\0&=l^2-5l-104\end{align*}

At this point, you can factor if that’s how you like to roll, or you can graph. To factor, simply figure out which pair of factors for 104 are 5 apart. 13 and 8 are, so you know that:

    \begin{align*}0&=(l-13)(l+8)\end{align*}

Since we’re dealing with physical lengths, we don’t need to worry about negative solutions. All we need is l=13.

To graph, simply plug the equation into your calculator. You’ll probably need to zoom out once.

In the image above, you see that the big parabola formed by y=x^2-5x-104 has a zero at x=13. That’s the value of l you’re looking for.

Test 6 calculator section #34 please!

I’ll give you two good ways to go on this one. First, the algebra. If you know the graph of y=3x^2-14x goes through the point (a,a), then you can substitute a in for x and y and solve!

    \begin{align*}a&=3a^2-14a\\0&=3a^2-15a\\0&=a^2-5a\\0&=a(a-5)\end{align*}

From that, you know a must equal 5, since the question distinguishes between (0,0) and (a,a).

The other way to go is to use your graphing calculator—just find the intersections between y=3x^2-14x and y=x.

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Hi! Can you please explain number 25 section 4 test 1?

Yep! You’re given this equation: h=-4.9t^2+25t. That’s the height of a ball t seconds after it is launched. You need to know how much time goes by before the ball hits the ground. The key thing to remember is that h will equal zero when the ball hits the ground, so you’re really solving 0=-4.9t^2+25t for t.

Since this is the calculator section, you have lots of options at your disposal.

First, you can backsolve. Just plug each answer choice in for t until you get something close to zero as an answer. This is a little tricky because you won’t get exactly zero, but the question does say “approximately.” Below is my calculator’s screen after I tried choice C, recognized that that was probably too high, remembered that a falling ball will get closer and closer to 0 the longer it’s in the air, and then tried D, which turned out to be pretty close to zero.

t1s4-25a

Another way to go is to graph. This is my favorite way for this question, because you can just eyeball the graph and see the right answer—you don’t even need to use the zero function to get it exact. Look:

Pretty obvious that that parabola hits y = 0 around x = 5, right?

The last way you can go is to factor and solve algebraically.

0=-4.9t^2+25t

0=t(-4.9t+25)

Since you aren’t interested in the fact that the height was zero at time zero, solve 0=-4.9t+25.

0=-4.9t+25

-25=-4.9t

5.102...=t