Solving Linear/Quadratic Functions: h=-16tsq.+40t+27

[Raye]

Help!!! I am taking a college level math course and am having trouble solving 2 problems. One problem is

h=-16tsq.+40t+27. Find the maximum height and the time when it reaches it.

For x of the vertex my answer is -5/4. For y of the vertex, I have two answers. One 32 seconds and another 39.5 seconds.

The second problem on solving

let f(x)=mx+b and let h(x)=f(f(x)). If the x- and y-intercepts of h(x) are 7 and -5 respectively, find m and b of f(x).

Thanks for your help.

 

[Deleted member 4993]

Re: Solving Linear/Quadratic Functions

Help!!! I am taking a college level math course and am having trouble solving 2 problems. One problem is h=-16tsq.+40t+27. Find the maximum height and the time when it reaches it.

For x of the vertex my answer is -5/4. For y of the vertex, I have two answers. One 32 seconds and another 39.5 seconds.

For a proper function - you cannot have multiple 'y's at a given 'x'. Please show work

The second problem on solving let f(x)=mx+b and let h(x)=f(f(x)).

If the x- and y-intercepts of h(x) are 7 and -5 respectively, find m and b of f(x). Thanks for your help.

Hint: Write the expression for h(x) - expanding f(f(x) - it will be linear in 'x'

 

[mmm4444bot]

h=-16tsq.+40t+27

We type this as: h = -16t^2 + 40t + 27

For x of the vertex <<<< I think you're trying to say, "the t-coordinate of the vertex point"

For y of the vertex, I have ... 32 seconds and another 39.5 seconds.

Again, it looks like you're confusing symbols. If y = h(t), then y is the height, not the time. The variable t represents the time.

Let's start with some notation.

You're given a function called h. Its independent variable is the symbol t (where t represents some number of elapsed seconds).

This function's dependent variable is h(t). The symbol h(t) represents the height of some object at any given value of t in the domain.

If you were to graph this function, after defining y = h(t), then you would graph it on a (t, y) coordinate system. In other words, the horizontal axis is called the t-axis.

You're probably used to dealing strictly with an x-axis and y-axis. You need to begin expanding your horizons because the axes are often not named x and y in applied math (i.e., word problems).

h(t) = -16t^2 + 40t + 27

You did not show your work for finding -5/4 as the t-coordinate of the vertex point, so I do not understand where the negative sign came from. (This value of t is WHEN the maximum height is reached; we cannot have a negative number of elapsed seconds.)

Please check your work for finding the time when the maximum height is reached.

You wrote that you found two y-coordinates for the vertex. You listed them as 32 seconds and 39.5 seconds. Again, you did not show your work, so I don't have any way of knowing what you did.

I can say that there is only ONE vertex in a quadratic function. I can say that the height at that point is NOT some number of seconds.

To find the y-coordinate of the vertex, you simply take the correct value for the t-coordinate of the vertex and evaluate h(t). You will get one number, and that is the maximum height.

If you would like more help with your first exercise, then please show your work.

On the second exercise, do you understand that f(f(x)) is a composite function? Do you know how to find its definition from f(x)?