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# Quadratic word problems (vertex form)

Given a quadratic function that models the height of an object being launched from a platform, we analyze the function to answer questions like "what is the height of the platform?" or "when does the object reach its maximum height?".

## Want to join the conversation?

- What if I need to solve for the constant a using vertex form, an then solve for f(x) when the point is 200 units away from the vertex? This was a question on my precalculus homework, and I've been struggling with this problem for the last two days. Any help is appeeciated!(19 votes)
- Around 700 days late but what do you mean by that? You want to know the y value when x is 200 units away? Find the vertex first and then add or subtract 200 units from the x value, then you can substitute and find out y after.(9 votes)

- why don't you solve the equation out at the second question in3:46? I don't understand why you brought the 180 over.(6 votes)
- Your question, I think, can be rephrased as, "How do we know that 180 is the maximum height of the launched object?" It's explained starting at about4:10. The not-so-short version is this:

The term -5(x-4)^2 will always evaluate as a non-positive number. This is because (x-4)^2 will always be evaluated as a positive value (because the square of anything is positive) and any positive multiplied by -5 will be negative. The greatest value of h(x) then is when (x-4)^2 = 0, or when x=4.

h(4) = -5(4-4)^2 + 180

h(4) = -5(0)^2 + 180

h(4) = -5(0) +180

h(4) = 0 + 180

h(4) = 180

Any different value for x results in a bigger (x-4)^2 value, results in a more negative -5(x-4)^2 value, and results in adding a more negative number to 180. So 180 is the highest h(x) will ever reach.(24 votes)

- Just a quick question: Is a straight line considered a parabola? Why?(3 votes)
- No, A parabola is set of all points which are an
**equal**distance away from a given point(the focus) and a given line(the directrix). (look up: focus and directrix of a parabola)

Parabola is sitting in between of the focus and the directrix.

now if you put line there instead of parabola(assume that it is a parabola), then take any point on that line and it should be equal distance away from the focus and from the directrix.

But you'll notice that because line and directrix are parallel, distance between them will never change, but distance from a point to the focus changes. so equality won't be true for all points. and because of that a straight line is not considered a parabola.(9 votes)

- Why did Sal ask how high is the platform, instead of asking how high is the object?

Isn't the height of the platform going to be zero?(3 votes)- The height of the platform isn't zero because it isn't the ground. A platform is higher than the ground. So since the ground is zero, the platform is higher than zero.(9 votes)

- What would you solve the maximum height question if the -5 were actually a positive? If it were you couldn´t assume that 280 would be the maximum positive height. Or is it impossible for it to be positive?(3 votes)
- Your assumption at the end is correct, it cannot be positive because in these types of problems, the coefficient of the x^2 term has to do with the acceleration due to the pull of gravity, so this will always be negative (trying to get the object to fall to the ground). It is common in these types of launching problems to see -9.8 as coefficient of x^2.(8 votes)

- I have a question regarding1:26. Generally speaking, if the leading coefficient is positive, will the parabola be facing up? Similarly, if the coefficient is negative (as in this case) will the parabola be facing down?(3 votes)
- Yep, exactly right. Just be careful any distributing works out to get the right sign.(3 votes)

- couldn't I just make a table to find the vertex?(3 votes)
- yes, but where do you know where to start or find values you want? You could have a table with 10 or more points before finding the vertex. If there is an easier way, why go through more effort?(2 votes)

- Since vertex hasn't been covered in my class, I don't get it. Someone please explain it to me?(2 votes)
- It's a unique way to view quadratic equations, in terms of the vertex. As you may know, a parabola looks like a bent "U" or "∩" shape.

There's a point on the parabola where it stops before curving back again in the opposite direction. This is known as the**vertex**.

The most simple quadratic function,`f(x) = x^2`

, has a vertex of`(0, 0)`

.

You can*transform*the function so that the position of the vertex can be changed. If you haven't learned function transformations yet, don't worry about it for now.*Translating*the function,`f(x) = (x - h)^2 + k`

has a vertex at`(h, k)`

. This is nearly the vertex form.

The entire vertex form of a quadratic equation is:`f(x) = A(x - h)^2 + k`

. While`A`

doesn't change where the vertex is, it determines how sharply the parabola curves and the direction it faces. If`A > 0`

, the parabola is U. If`A < 0`

, the parabola is ∩.

But this is similar to how in the standard form,`f(x) = ax^2 + bx + c`

, when`a > 0`

the parabola is U, and when`a < 0`

the parabola is ∩.

An example of a quadratic equation in vertex form is:`y = 2(x + 2)^2 + 4`

; the vertex is located at`(-2, 4)`

.

A few things about vertex form calculations:

- You can**convert**any quadratic function from standard form to vertex form and vice versa.

- If you know the vertex point of a parabola, and`1`

other point, you can find the**entire equation**of the parabola in vertex form.

There is content later on in this Unit (Unit 14) in the Algebra 1 course - you can find content on the vertex form there.

Hope this helps for later on; it was challenging to write.(2 votes)

- At about2:24minutes into the video, shouldn't the question be; Whats the height of the platform + the height of the object?

The function is defined for this right and not just for the platform?(2 votes) - How would you solve how long it would take for the object to hit the ground?(1 vote)
- You have to have a quadratic equation, and then you find the solution (x-intercept). This will help tell how long it takes to hit the ground. Unless otherwise indicated, the assumption is that at time 0, the object was launched along a quadratic part, so the positive x value of the x intercept should be the time it took.(2 votes)

## Video transcript

- [Sal] An object is
launched from a platform. Its height in meters, x
seconds after the launch, is modeled by: h of x is
equal to negative five times x minus four squared plus 180. Normally, when they talk
about seconds or time, they usually would use the variable t, but we can roll with x being that. Let's think about what's
going to happen here. Lemme just visualize it. Lemme draw an h axis for our height. Let me draw an x axis. An x axis. At time x is equal to zero. We're on a platform, so we're already gonna have some height. At time x is equal to zero. I'm used to saying time t equals zero, but at time x is equal to zero, we're already gonna have some height 'cause we're on some platform. And then we're gonna
launch this projectile. And, it's gonna go in
the shape of a parabola, and it's gonna be a
downward-opening parabola. You might say: "Sal, how do you know "it's gonna be a
downward-opening parabola?" Gonna look something like that. I didn't draw it exactly perfectly, but you get, hopefully, the point. The reason why I knew it was a parabola, in particular a downward-opening parabola, is when you look at what's going on here. This is written in vertex
form but it's a quadratic. In vertex form, you have an
expression with x squared, and then you're multiplying by negative five right over here. This tells us that it's
gonna be downward-opening. If you were to multiply this out, x minus four squared is gonna be x squared plus something else plus something else, then you're gonna have to multiply all those terms by negative five, your leading term is gonna
be negative five x squared. Once again, it's gonna be
a downward-opening parabola that looks something like that. So, given this visual
intuition that we have, let's see if we can answer
some questions about it. The first one I'd like to answer is how high is the platform. How high is the platform? How high is the platform? I encourage you to pause the video, and try to figure that out. What is that value right over there? Well, as you can see, we are at that value
at time x equals zero. So to figure out how high is the platform, we essentially just have
to evaluate h of zero. That's going to be negative five times negative four squared plus 180. I just substituted x with zero. Negative four squared is 16. Negative five times 16 is negative 80. Plus 180. So this is going to be equal to 100. So the platform is 100 meters tall. Remember, the height is given in meters. Now, the next question I have is, how many seconds after launch
do we hit our maximum height? So our maximum height,
if we're talking about a downward-opening parabola,
it's going to be our vertex, is going to be our maximum height. And so, the x value of that would tell us how long after takeoff,
how long after, or launch, do we hit the maximum height. Trying to use a color you can see. What is this x value right over here? Once again, pause the video, and see if you can figure it out. We're trying to answer how long after launch is the maximum height. Well, it's going to be the
x coordinate of our vertex. How do we figure that out? Well, this quadratic has
actually been written already in vertex form, which
makes it sound like it should be relatively easy to figure
out the vertex over here. To appreciate that, we
have to see the structure in the expression, is one
way to think about it. Let's think about what's going on. You have this 180. And then you this other
term right over here. Anything squared is gonna be nonnegative. So x minus four squared is
always gonna be nonnegative. But then you always multiply
that times a negative five, so this whole thing is
gonna be non-positive. So, it will never add to the 180. Your maximum value is when
this term right over here is going to be equal to zero. And when is this term
going to be equal to zero? In order to make this term equal to zero, then x minus four needs
to be equal to zero. The only way to get x minus
four to be equal to zero is if x is equal to four. Just by looking at this, you say: "Hey, what makes this zero?" Four. X equals four will make this zero. This is right over there. If I were to write h of
four, this is going to be, this term is gonna go to zero, and you're gonna be left with the 180. There you go, this right over here. The maximum height is 180. It happens four seconds after launch. Now, the last question I'll ask you is, how long after launch do
we get to a height of zero? So, for what x makes our height zero? To do that, we have to solve
h of x is equal to zero. Or, we can write h of x as negative five times x minus four squared
plus 180 is equal to zero. And, once again, pause the video, and see if you can solve this. Could subtract 180 from both sides. You get negative five
times x minus four squared, is equal to negative 180. We can divide both sides by negative five. We get x minus four
squared is equal to 36. Scroll down a little bit. Then, we could take the plus
and minus square root, I guess you could say. And so, that will give us x minus four could be equal to six. Or, x minus four is equal to negative six. In this first situation,
add four to both sides, you get x is equal to 10. Or, you add four to both sides here, you get x is equal to negative two. Now, we're dealing with time here, so negative two would've been in the past if it wasn't sitting on the platform and if it was just
continuing its trajectory, I guess you could say, backwards in time. But that's not the x that we
wanna take into consideration. We want the positive time value,
and that's right over here. That is when x is equal to 10. 10 seconds after takeoff, our height is going to be equal to zero. If the ground is at height of zero, if it's at sea level, I guess, then, that's when our projectile
is going to hit the ground.