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I need help with College Algebra

B

BlancoNino

Diamond Member
I'm studying for a math test tomorrow and I'm having trouble with the height equation. The question asks:

A frog leaps from a stump 3 feet high and lands 4 feet from the base of the stump. We can consider the initial position of the frog to be at (0, 3) and its landing position to be at (4, 0).

It is determined that the height h of the frog as a function of its distance x from the base of the stump is given by:

h(x) = -.5x^2 + 1.25x + 3

a) how high was the frog when its horizontal distance x from the base of the stump was 2 feet?

B) What was the horizontal distance from the base of the stump when the frog was 3.25 feet above the ground?

c) At what horizontal distance from the base of the stump did the frog reach its highest point?

d) What was the maximum height reached by the frog?

First of all I don't understand the function given above. Besides the 3, where did the numbers come from in the function given above?

I'm totally lost in this section. This isn't my homework and we're not graded on this, but if I can understand this problem I can understand about 15% of the test tomorrow. Thanks a million.
 
Originally posted by: BlancoNino
I'm studying for a math test tomorrow and I'm having trouble with the height equation. The question asks:

A frog leaps from a stump 3 feet high and lands 4 feet from the base of the stump. We can consider the initial position of the frog to be at (0, 3) and its landing position to be at (4, 0).

It is determined that the height h of the frog as a function of its distance x from the base of the stump is given by:

h(x) = -.5x^2 + 1.25x + 3

a) how high was the frog when its horizontal distance x from the base of the stump was 2 feet?

b) What was the horizontal distance from the base of the stump did the frog reach its highest point?

c) At what horizontal distance from the base of the stump did the frog reach its highest point?

d) What was the maximum height reached by the frog?

First of all I don't understand the function given above. Besides the 3, where did the numbers come from in the function given above?

I'm totally lost in this section. This isn't my homework and we're not graded on this, but if I can understand this problem I can understand about 15% of the test tomorrow. Thanks a million.
Click to expand...

1. the h(x) function solves for the height as a function of the horizontal distance (x). plug the horizontal distance (x) in to find the height.

you can "plug and chug" different values of x (0-4) into the function to find all the answers also...
 
Originally posted by: BlancoNino
I'm studying for a math test tomorrow and I'm having trouble with the height equation. The question asks:

A frog leaps from a stump 3 feet high and lands 4 feet from the base of the stump. We can consider the initial position of the frog to be at (0, 3) and its landing position to be at (4, 0).

It is determined that the height h of the frog as a function of its distance x from the base of the stump is given by:

h(x) = -.5x^2 + 1.25x + 3

a) how high was the frog when its horizontal distance x from the base of the stump was 2 feet?

b) What was the horizontal distance from the base of the stump did the frog reach its highest point?

c) At what horizontal distance from the base of the stump did the frog reach its highest point?

d) What was the maximum height reached by the frog?

First of all I don't understand the function given above. Besides the 3, where did the numbers come from in the function given above?

I'm totally lost in this section. This isn't my homework and we're not graded on this, but if I can understand this problem I can understand about 15% of the test tomorrow. Thanks a million.
Click to expand...

I"m loath to do homework for people, but I was curious to see whether I remembered any of this stuff. This may be completely wrong:

A. Put x=2 into equation.
Y = -.5x^2 + 1.25x + 3
Y = -.5(2^2) + 1.25(2) + 3
Y = -2 + 2.5 + 3
Y = 3.5

B. Graph the equation Y = -.5x^2 + 1.25x + 3. Find what X is where Y is at the highest point. The graph should be an inverted parabola, so should have a clear highest point (plug the highest value of Y into the equation and solve for X).

C. Isn't this the same as B?

D. Graph the equation Y = -.5x^2 + 1.25x + 3. Find out what Y is at the graph's highest point.
 
a) Evaluate h(2).
b) Find h'(x) and solve for zero.
c) See above?
d) After solving b), you should have a value for x for which the derivative of the function h is zero. Plug that into h, and you have your answer.

No graphics calculator required. You should have learned this stuff in grade 11 maths (or, at least, that's what the kids down here in Australia do).
 
Originally posted by: Sc4freak
a) Evaluate h(2).
b) Find h'(x) and solve for zero.
c) See above?
d) After solving b), you should have a value for x for which the derivative of the function h is zero. Plug that into h, and you have your answer.

No graphics calculator required. You should have learned this stuff in grade 11 maths (or, at least, that's what the kids down here in Australia do).
Click to expand...

We do here too, but the problem is that I haven't been in 11th grade for 7 years and the math classes we pass in highschool count as jack shit for credits when we're in college because they want it taught differently. At least around here that's how it works.
 
So the function given isn't something I can figure out on my own and must be provided by the question for me to solve? Cause that's where I'm mostly confused.
 
Yep. The function describes the height of the frog with respect to its distance. You can't answer the question unless they give you the function (or give you a way to derive it).
 
Originally posted by: Sc4freak
Yep. The function describes the height of the frog with respect to its distance. You can't answer the question unless they give you the function (or give you a way to derive it).
Click to expand...

It may be because I'm super tired, but I still don't understand part B. You said solve for zero. Quadratic formula?
 
Originally posted by: BlancoNino
Originally posted by: Sc4freak
Yep. The function describes the height of the frog with respect to its distance. You can't answer the question unless they give you the function (or give you a way to derive it).
Click to expand...

It may be because I'm super tired, but I still don't understand part B. You said solve for zero. Quadratic formula?
Click to expand...
He was saying to take the derivative and solve for zero, which I'm guessing you haven't learned in algebra. I'd recommend just graphing it.
 
No, you don't need the quadratic formula. h'(x) means the derivative of h(x). Finding the derivative of a polynomial is easy, you just need to drop the powers by 1 and multiply.

eg. f(x) = 5x^3 + 2x^2 + 9x + 10
f'(x) = 15x^2 + 4x + 9

So h'(x) = -x + 1.25. Make h'(x) equal to zero, solve for x, and you get 1.25.
 
Originally posted by: Sc4freak
No, you don't need the quadratic formula. h'(x) means the derivative of h(x). Finding the derivative of a polynomial is easy, you just need to drop the powers by 1 and multiply.

eg. f(x) = 5x^3 + 2x^2 + 9x + 10
f'(x) = 15x^2 + 4x + 9

So h'(x) = -x + 1.25. Make h'(x) equal to zero, solve for x, and you get 1.25.
Click to expand...

College ALGEBRA.

The vertex of the parabola has an x-coord of x=-b/2a if the equation has the form y=ax^2+bx+c


And I know why x=-b/2a, so don't try getting coy with me ATOT. I hardly think his algebra teacher wants him using differential calculus.
 
Wow, you guys are WAY overcomplicating this. Since the function is a second degree function, that means it has at most one min or max. You can therefore find the x coordinate of the min/max by using the formula x=(-b)/2a. Remember that from way back in precalculus guys? No need to throw calculus into this problem!

Now that you have the x value of the highest point, plug it into the equation and get the Y value. There is your highest point.
 
Originally posted by: Brigandier
Originally posted by: Sc4freak
No, you don't need the quadratic formula. h'(x) means the derivative of h(x). Finding the derivative of a polynomial is easy, you just need to drop the powers by 1 and multiply.

eg. f(x) = 5x^3 + 2x^2 + 9x + 10
f'(x) = 15x^2 + 4x + 9

So h'(x) = -x + 1.25. Make h'(x) equal to zero, solve for x, and you get 1.25.
Click to expand...

College ALGEBRA.
Click to expand...

Heh, whoops. Guess I must've skipped over that part.
 
Originally posted by: Hacp
Wow, you guys are WAY overcomplicating this. Since the function is a second degree function, that means it has at most one min or max. You can therefore find the x coordinate of the min/max by using the formula x=(-b)/2a. Remember that from way back in precalculus guys? No need to throw calculus into this problem!

Now that you have the x value of the highest point, plug it into the equation and get the Y value. There is your highest point.
Click to expand...

I'm still stuck on b lol. What was the horizontal distance from the base of the stump when the frog was 3.25 feet above the ground?

You guys are all helping. And I appreciate it. I've been sick all week and so I'm forced to cram tonight.
 
Originally posted by: BlancoNino
Originally posted by: Hacp
Wow, you guys are WAY overcomplicating this. Since the function is a second degree function, that means it has at most one min or max. You can therefore find the x coordinate of the min/max by using the formula x=(-b)/2a. Remember that from way back in precalculus guys? No need to throw calculus into this problem!

Now that you have the x value of the highest point, plug it into the equation and get the Y value. There is your highest point.
Click to expand...

I'm still stuck on b lol. What was the horizontal distance from the base of the stump when the frog was 3.25 feet above the ground?

You guys are all helping. And I appreciate it. I've been sick all week and so I'm forced to cram tonight.
Click to expand...

Substitute 3.25 feet for the y value in your equation and solve.

Y=-.5x^2 + 1.25x + 3

3.25=-.5x^2 + 1.25x + 3
 
Originally posted by: BlancoNino
BTW there should be 2 answers for B since the frog is jumping from 3 into the air and then coming back down.
Click to expand...

Is 3.25 the highest point? If it is, there is only one answer. If it isn't, there is a possibility for two answers.
 
a) how high was the frog when its horizontal distance x from the base of the stump was 2 feet?

b) What was the horizontal distance from the base of the stump did the frog reach its highest point?

c) At what horizontal distance from the base of the stump did the frog reach its highest point?

d) What was the maximum height reached by the frog?
Click to expand...

That was the list of questions you gave. You said you were stuck on question B.
 
Shit, I messed up typing that.

B) What was the horizontal distance from the base of the stump when the frog was 3.25 feet above the ground?
 
3.25 is the y coordinate. Your original equation is
h(x) = -.5x^2 + 1.25x + 3. Since Y is basically h(x), you sub 3.25 for h(x) and get
3.25=-.5x^2 + 1.25x + 3

Solve that equation using the quadratic formula. Hope you know how to do that.
 
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