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Algebra 1

Course: Algebra 1>Unit 14

Lesson 4: 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?".

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• 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!
• 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.
• why don't you solve the equation out at the second question in ? I don't understand why you brought the 180 over.
• 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 about . 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.
• 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?
• 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.
• 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?
• 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.
• Just a quick question: Is a straight line considered a parabola? Why?
• 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.
• I have a question regarding . 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?
• Yep, exactly right. Just be careful any distributing works out to get the right sign.
• couldn't I just make a table to find the vertex?
• 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?
• Sal says that at x seconds the object is going to have some height, is that the height of the object?