Need help with a past exam paper and this txt book please

[kingsala]

John likes to play baseball with his friends in the local playing fields. Last
week John hit his best shot ever. The trajectory of the baseball after he
hit it can be modeled by the quadratic equation
y= -x2/40 + 31x/ 40 + 4/5

where y represents the height in metres of the baseball above the ground,
and x represents the horizontal distance in metres of the baseball from the
position where it was struck by John. Assume that the surface of the
playing field is horizontal.
(a) The graph of y= -x2/40 + 31x/ 40 + 4/5 is a parabola.

(i) Is the parabola u-shaped or n-shaped? How can you tell this from
the equation? [1]
(ii) Use algebra to find the x-intercepts and y-intercept.
(iii) Find the equation of the axis of symmetry, explaining your
method. Use this information to find the coordinates of the vertex, giving your answers to two decimal places.
(iv) Provide a sketch of the graph of the parabola, either by hand or by using Graphplotter.

(b) In this part of the question, you are asked to consider the trajectory
of the baseball modeled by the equation y= -x2/40 + 31x/ 40 + 4/5 in conjunction with the results that you found in part (a).

(i) Find the height of the baseball when it was 5 metres horizontally from the position where John hit it.
(ii) Use your answer to part (a)(iii) to find the maximum height reached by the baseball.
(iii) What does the y-intercept represent in the context of this model?
(iv) Assuming that the baseball was not caught, how far was it horizontally from where John hit it when it first landed on the ground? Explain your answer.

 

[MarkFL]

Hello and welcome to MHB! :D

Unfortunately, the trajectory of the ball as you have stated it does not make sense, at least to me. Can you write it out so that we know what it actually is?

 

[kingsala]

Please help , not got long to get my head round this

John likes to play baseball with his friends in the local playing fields. Last
week John hit his best shot ever. The trajectory of the baseball after he
hit it can be modeled by the quadratic equation
y= -x2/40 + 31x/ 40 + 4/5

where y represents the height in metres of the baseball above the ground,
and x represents the horizontal distance in metres of the baseball from the
position where it was struck by John. Assume that the surface of the
playing field is horizontal.
(a) The graph of y= -x2/40 + 31x/ 40 + 4/5 is a parabola.

(i) Is the parabola u-shaped or n-shaped? How can you tell this from
the equation? [1]
(ii) Use algebra to find the x-intercepts and y-intercept.
(iii) Find the equation of the axis of symmetry, explaining your
method. Use this information to find the coordinates of the vertex, giving your answers to two decimal places.
(iv) Provide a sketch of the graph of the parabola, either by hand or by using Graphplotter.

(b) In this part of the question, you are asked to consider the trajectory
of the baseball modeled by the equation y= -x2/40 + 31x/ 40 + 4/5 in conjunction with the results that you found in part (a).

(i) Find the height of the baseball when it was 5 metres horizontally from the position where John hit it.
(ii) Use your answer to part (a)(iii) to find the maximum height reached by the baseball.
(iii) What does the y-intercept represent in the context of this model?
(iv) Assuming that the baseball was not caught, how far was it horizontally from where John hit it when it first landed on the ground? Explain your answer.

 

[MarkFL]

Since edits don't show up as new posts, I did not realize you have edited your post. I apologize for this delay. When you make major changes like that, it is best to do so in a new post, that way we'll know there is additional information to view. :D

So, the trajectory of the ball is given by:

\(\displaystyle y=-\frac{x^2}{40}+\frac{31x}{40}+\frac{4}{5}\)

Let's take the questions one at a time. We are first asked if the parabola given opens up or down. We can tell from the coefficient of the leading or squared term. If it is negative, then we know it opens down and if it is positive then we know it opens up.

Which do we have here?

 

[kingsala]

Since edits don't show up as new posts, I did not realize you have edited your post. I apologize for this delay. When you make major changes like that, it is best to do so in a new post, that way we'll know there is additional information to view. :D

So, the trajectory of the ball is given by:

\(\displaystyle y=-\frac{x^2}{40}+\frac{31x}{40}+\frac{4}{5}\)

Let's take the questions one at a time. We are first asked if the parabola given opens up or down. We can tell from the coefficient of the leading or squared term. If it is negative, then we know it opens down and if it is positive then we know it opens up.

Which do we have here?

The parabola open up

 

[MarkFL]

What is the coefficient of the leading term?

 

[kingsala]

This is the parts I just don't understand :(

If I am right its the leading term is the term with the highest power of x

 

[MarkFL]

Yes, when a polynomial is arranged such that the exponents decrease from left to right, then the leading term will be the term having the greatest exponent on the variable. In this case, the coefficient on the leading term is:

\(\displaystyle -\frac{1}{40}\)

and because it is negative, then we know the parabola opens down. Think about throwing a ball, wouldn't you expect its trajectory to open down rather than up?

 

[Deveno]

We can check this another way:

Let's pick 3 points $x_1 < x_2 < x_3$.

If $f(x_2) < f(x_1),f(x_3)$ it opens "up", if:

$f(x_2) > f(x_1),f(x_3)$ it opens "down".

Of course, to do this properly, we need to pick $x_2$ near the vertex and $x_1,x_3$ well away from it, or else we may not find out anything useful.

Since we need to find the axis of symmetry anyway, maybe we should address this now: do you know how to "complete the square"?

Another strategy for picking our 3 x's suggests itself here: take $x_1 = 0$, take $x_3 = 31$ (as the ball is then at the same height at which it started from) and take $x_2$ to be the midpoint between them.

 

[MarkFL]

For part ii) finding the intercepts...let $x=0$ and then resulting value of $y$ gives you the $y$ intercept. Let $y=0$ and solve for $x$ to get the $y$-intercepts.

Then you have several options for finding the axis of symmetry...and since you already have the roots at this stage, just take the vertical line midway between them.

 

[kingsala]

For part ii) finding the intercepts...let $x=0$ and then resulting value of $y$ gives you the $y$ intercept. Let $y=0$ and solve for $x$ to get the $y$-intercepts.

Then you have several options for finding the axis of symmetry...and since you already have the roots at this stage, just take the vertical line midway between them.

I understand the open and closed side of things but I we were working from question one (i) onwards what is the explanations of them at least the I can work through them at my own pace and hopefully have the right answers by the end of the week.#

Thanks again guys for the help, Any more help would be great

 

[MarkFL]

I would recommend posting what you have so far, up to the first point you are stuck, and then we can help get you unstuck and proceed from there. (Nod)

 

[nicholar1]

Hi,

I was wondering if someone could explain how to do part b(ii) of this question. How would I find the maximum height?

/Nichola

 

[Fermat1]

Hi,

I was wondering if someone could explain how to do part b(ii) of this question. How would I find the maximum height?

/Nichola

Here's the way suggested by the question. Going back, you want to find the vertex co-ordinates. This will be when the ball is halfway on it's journey. So find the total distance the ball travels by setting $y=0$ (the ball will be on the ground) and solving for $x$. The equation $x^2-31x-3=0$ has 2 solutions, namely $x=-1,32$. Reject the negative root to conclude it travels 32m. So the vertex co-ordinates are (16,34/5). So the maximum height is $34/5$m