Intercepts

[Susan Smith]

Hi everyone my friend stocked with this question and don't know how to do it
can you guys help me with it?
it would be great to show me how to do it so I can explain it to her.

 

[MarkFL]

Hello, and welcome to FMH!

(i) To find the \(y\)-intercept, let \(x=0\) and evaluate the given function...what do you get?

 

[Susan Smith]

Hello, and welcome to FMH!

(i) To find the \(y\)-intercept, let \(x=0\) and evaluate the given function...what do you get?

to be honest, I am not good at intercept if you can help me with answers it would be helpful.

 

[MarkFL]

to be honest, I am not good at intercept if you can help me with answers it would be helpful.

Simply providing the answers would be minimally helpful. Helping you understand how to find the answers yourself would be much more helpful. How will you be able to explain this to your friend if you don't know where the answers come from?

We're trying to find the point at which the given function crosses the \(y\)-axis. This axis is the line \(x=0\). So, if we let \(x=0\) in the given function, then the resulting value for \(y\) will tell us where on the \(y\)-axis the function is.

[MATH]y(0)=-0.23(0)^2+1.87(0)+1.5=?[/MATH]

 

[Susan Smith]

Simply providing the answers would be minimally helpful. Helping you understand how to find the answers yourself would be much more helpful. How will you be able to explain this to your friend if you don't know where the answers come from?

We're trying to find the point at which the given function crosses the \(y\)-axis. This axis is the line \(x=0\). So, if we let \(x=0\) in the given function, then the resulting value for \(y\) will tell us where on the \(y\)-axis the function is.

[MATH]y(0)=-0.23(0)^2+1.87(0)+1.5=?[/MATH]

so is gonna be 1.5 ?
or I need to add them together ?

 

[MarkFL]

so is gonna be 1.5 ?
or I need to add them together ?

Yes, the two terms with zero as a factor will be zero, and so we are left with 1.5.

[MATH]y(0)=1.5[/MATH]
So, we know the point \((0,1.5)\) is on the curve.

For the next question, you are being asked to find \(y(3)\)...what do you find?

 

[Susan Smith]

Yes, the two terms with zero as a factor will be zero, and so we are left with 1.5.

[MATH]y(0)=1.5[/MATH]
So, we know the point \((0,1.5)\) is on the curve.

For the next question, you are being asked to find \(y(3)\)...what do you find?

this one is a bit confusing can you explain more, please?
for the second question that said subtract x=3

 

[MarkFL]

We are given that:

[MATH]y(x)=-0.23x^2+1.87x+1.5[/MATH]
For this second question, we are told to let \(x=3\) and so:

[MATH]y(3)=-0.23(3)^2+1.87(3)+1.5=?[/MATH]

 

[Susan Smith]

We are given that:

[MATH]y(x)=-0.23x^2+1.87x+1.5[/MATH]
For this second question, we are told to let \(x=3\) and so:

[MATH]y(3)=-0.23(3)^2+1.87(3)+1.5=?[/MATH]

it will be 5.73. right ?

 

[MarkFL]

it will be 5.73. right ?

I get a different value...can you show your work?

 

[Susan Smith]

I get a different value...can you show your work?

it was my bad. it would be 9.18
and i think it will be like this
0.23x9+1.87x3+1.5=9.18

right ?

 

[MarkFL]

it was my bad. it would be 9.18
and i think it will be like this
0.23x9+1.87x3+1.5=9.18

right ?

You're neglecting the negative sign in front of the leading coefficient, but otherwise you have the right idea. Taking that into account, what do you get?

Note: Please don't use an "x" to denote multiplication, as this can be confused with the independent variable.

 

[Susan Smith]

You're neglecting the negative sign in front of the leading coefficient, but otherwise you have the right idea. Taking that into account, what do you get?

Note: Please don't use an "x" to denote multiplication, as this can be confused with the independent variable.

alright sorry about the x and for the negative sign it going to be 5.04?
I am really sorry for being confused

 

[MarkFL]

alright sorry about the x and for the negative sign it going to be 5.04?
I am really sorry for being confused

Yes, that's what I get as well. So, given that, will the missile be able to clear the obstacle?

 

[Susan Smith]

Yes, that's what I get as well. So, given that, will the missile be able to clear the obstacle?

no, I think because it's 5.04 and it is higher right?

 

[MarkFL]

If the obstacle is 4 cm high, and the missile is 5.04 cm above the ground when it reaches the obstacle, then since 5.04 > 4, I would say the missile will clear the obstacle.

 

[Susan Smith]

If the obstacle is 4 cm high, and the missile is 5.04 cm above the ground when it reaches the obstacle, then since 5.04 > 4, I would say the missile will clear the obstacle.

aw great
I understand it now thanks
now we get to the
third question
it would be great to help me with that as well

 

[MarkFL]

To find the \(x\)-intercepts, we want to let \(y=0\), and then solve for \(x\), but keep in mind that the implied domain here is:

[MATH][0,\infty)[/MATH]
or:

[MATH]0\le x[/MATH]
What do you find?

 

[Susan Smith]

To find the \(x\)-intercepts, we want to let \(y=0\), and then solve for \(x\), but keep in mind that the implied domain here is:

[MATH][0,\infty)[/MATH]
or:

[MATH]0\le x[/MATH]
What do you find?

I did not understand sorry can you explain it easier?

 

[MarkFL]

In the first part of the last question, we are simply being asked to find the \(x\)-intercepts. To do so, we let \(y=0\):

[MATH]-0.23x^2+1.87x+1.5=0[/MATH]
And then solve for \(x\). I'd begin by multiplying through by -100, so clear the decimals and get rid of the leading negative sign:

[MATH]23x^2-187x-150=0[/MATH]
Next, apply the quadratic formula...what do you get?