Question

A Labrador leaps over a hurdle. The function f(t) represents the height of the Labrador above the ground, in inches, at t seconds:

f(t) = -16t2 + 20t

Answer 1

A foxhound jumps over the same hurdle. The table shows the height of the foxhound above the ground g(t), in inches, at t seconds:

Time (t) g(t)
0 0
0.4 7.04
0.6 8.64
0.75 9
1.0 8
1.5 0


Part A: Compare and interpret the maximum of f(t) and g(t)? (4 points)

Part B: Which function has a greater x-intercept? What do the x-intercepts of the graphs of f(t) and g(t) represent? (4 points)

Part C: Determine the y-intercepts of both functions, and explain what this means in the context of the problem. (2 points)

Answer 2

@amistre64 @iGreen @ganeshie8

Answer 3

@Preetha @Michele_Laino @nincompoop

Answer 4

i got part A The maximum for t is 1.5 and the maximum for g(t) is 9. What that means is the foxhound jumped 9inches above the ground and the maximum for t means how long it took him.

Answer 5

for the first function, namely f(t), we have to find the coordinates of the vertex of the parabola f(t)

Answer 6

the coordinate of the vertex are V:
\[\Large V = \left( { - \frac{b}{{2a}}, - \frac{{{b^2} - 4ac}}{{4a}}} \right)\]
where a = -16, b= 20,a nd c=0, since I'm referring to a generic parabola:
f(t)= a*t^2+ b*t+c

Answer 7

and*

Answer 8

okay so for part B: i think its the first equation

Answer 9

i think its the first equation and the x represents how much time it took for him to leave and ground and get back to the ground

Answer 10

we have to write the quadratic function, using the coordinates from your table

Answer 11

okay.

Answer 12

let's conjecture this function for g(t):
\[g\left( t \right) = a{t^2} + bt\]
I wrote that function, since, from your table, the point (0,0) has to check the unknown function g(t)

Answer 13

now, the point t=1, g=8 belongs to the graph of g(t), so we can write this equation:
\[a + b = 8\]
furthermore the point t=1.5=3/2, and g=0, belongs to the graph of g(t), so we can write:
\[a\frac{9}{4} + b\frac{3}{2} = 0\]

Answer 14

so, we have to solve this algebraic system:
\[\Large \left\{ \begin{gathered}
a + b = 8 \hfill \\
a\frac{9}{4} + b\frac{3}{2} = 0 \hfill \\
\end{gathered} \right.\]
what are a, and b?

Answer 15

do i find the common factor of 4 and 3 and then add them?

Answer 16

you have to find the least common multiple between 4 and 2

Answer 17

okay so its 4 and then 3 gos to 6 and 9+6 is 15

Answer 18

here are your steps:
\[a\frac{9}{4} + b\frac{3}{2} = \frac{{9a + 3b \times 2}}{4} = ...?\]

Answer 19

what is \[{3b \times 2}=...?\]

Answer 20

uhm idk do i multply 3x2?

Answer 21

yes!

Answer 22

6

Answer 23

ok! so we have:
\[a\frac{9}{4} + b\frac{3}{2} = \frac{{9a + 3b \times 2}}{4} = \frac{{9a + 6b}}{4}\]

Answer 24

so thats my quadratic equation

Answer 25

no, it is your second equation of the system, namely:
\[\left\{ \begin{gathered}
a + b = 8 \hfill \\
\frac{{9a + 6b}}{4} = 0 \hfill \\
\end{gathered} \right.\]

Answer 26

or:\[\left\{ \begin{gathered}
a + b = 8 \hfill \\
9a + 6b = 0 \hfill \\
\end{gathered} \right.\]

Answer 27

okay.

Answer 28

now I solve the first equation for b, and I get:
b= 8 -a
then I substitute into the first equation, so I get:
\[\Large 9a + 6\left( {8 - a} \right) = 0\]

Answer 29

please solve it for a, what do you get?

Answer 30

uhm do you subtract 6 from both sides and get 9a(8-a)=-6 then what

Answer 31

no, you have to compute the multiplication between 6 and 8-a

Answer 32

okay so 6 times 8 and then its 6a so 9a(48-6a)=0

Answer 33

hit:
\[9a + 6\left( {8 - a} \right) = 9a + 48 - 6a\]

Answer 34

hint*

Answer 35

ok! so we get this equation:
\[9a + 48 - 6a = 0\]
what is a?

Answer 36

so now you have 3a+48=0 and subtract 48 to both sides and divide by 3 and get a=-16

Answer 37

ok!
now we have to substitute that value, into the first equation, and then we have to find the value of b, like this way:
\[ - 16 + b = 8\]
what is b?

Answer 38

at 16 to both sides and you get b=24

Answer 39

perfect, so the equation of g(t), is:
\[\Large g\left( t \right) = - 16{t^2} + 24t\]

Answer 40

okay

Answer 41

now what about answering part B?

Answer 42

the x-intercept of f(t), is given solving the subsequent equation:
\[\Large - 16{t^2} + 20t = 0\]
what is the non null root of that equation?

Answer 43

what does that mean

Answer 44

I factor out 4t and I get:
\[4t\left( { - 4t + 5} \right) = 0\]

Answer 45

now I apply the product canceling law, and I get these roots:
t=0, and t=5/4, so the x-intercept of f(t) is t= 5/4

Answer 46

alright

Answer 47

now we have to do the same using g(t), namely:
\[ - 16{t^2} + 24t = 0\]
what are the roots of that equation?

Answer 48

you could factor our 8t?

Answer 49

that's right!

Answer 50

\[8t\left( { - 2t + 3} \right) = 0\]

Answer 51

t=-4 and t=3/8? or 2.67?

Answer 52

using the product canceling law, I get:
t=0 and t= 3/2

Answer 53

ohhh

Answer 54

so the x-intercept of g(t) is 3/2, whereas the x-intercept of f(t) is 5/4

Answer 55

is 3/2 grater than 5/4, or is 5/4 greater than 3/2 ?

Answer 56

i think 3/2 is greater

Answer 57

ok! since:
\[\frac{3}{2} = \frac{6}{4} > \frac{5}{4}\]

Answer 58

ohh 5/4

Answer 59

so the x-intercept of g(t) is greater that the x-intercept of f(t)

Answer 60

so they represent time of how long he was off the ground right?

Answer 61

that's right!

Answer 62

Thanks!

Answer 63

Now what about part C?

Answer 64

the y-intercepts for both functions is y=0
ince we have the subsequent draw: