Graphing functions

[rachelmaddie]

Better. But where does your graph cut the y-axis. (Hint. On the y-axis, x=0 so what is y?)

y = 0?

 

[Harry_the_cat]

No. Put x=0 into your equation and work out y

 

[rachelmaddie]

No. Put x=0 into your equation and work out y

Can you help me with that? I don’t understand.

 

[Harry_the_cat]

Ok. Your equation is \(\displaystyle y=x^2+2x-3\).
That is the rule for working out y when you know x.
On the y-axis, x=0. (Every point on the y-axis has an x-coordinate of 0, right?)
So, if x=0, \(\displaystyle y=0^2+2*0-3=-3\)
So, (0, -3) is the y-intercept.

Now have a look at the last graph you drew. It should cut the y-axis at (0, -3). It doesn't quite, does it?

 

[rachelmaddie]

Ok. Your equation is \(\displaystyle y=x^2+2x-3\).
That is the rule for working out y when you know x.
On the y-axis, x=0. (Every point on the y-axis has an x-coordinate of 0, right?)
So, if x=0, \(\displaystyle y=0^2+2*0-3=-3\)
So, (0, -3) is the y-intercept.

Now have a look at the last graph you drew. It should cut the y-axis at (0, -3). It doesn't quite, does it?

Are you saying where I drew the line it is off?

 

[Harry_the_cat]

Your graph doesn't cut the y-axis at (0, -3).

 

[rachelmaddie]

View attachment 20502

Your graph doesn't cut the y-axis at (0, -3).

Right here where Ive circled?

 

[Harry_the_cat]

That's off too!
But I'm talking about where the graph cuts the Y axis, the vertical axis.

 

[rachelmaddie]

That's off too!
But I'm talking about where the graph cuts the Y axis, the vertical axis.

You mean the double red line that was from this other graph?

 

[Harry_the_cat]

I'm talking about where your parabola cuts the Y axis. It should pass through (0, -3). Look inside the circle. It doesn't pass through (0,-3).

 

[rachelmaddie]

View attachment 20505
I'm talking about where your parabola cuts the Y axis. It should pass through (0, -3). Look inside the circle. It doesn't pass through (0,-3).

Is this better?

 

[Harry_the_cat]

Is this better?View attachment 20506

Yes