If g(t)=t^2-4, then the graph of g(t) crosses the x-axis when t equals? @whpalmer4

Question

whpalmer4 · Answer

Do you know the equation for the line that is the x-axis?

anonymous · Answer

what do you mean?

whpalmer4 · Answer

the x-axis is a line, right?

anonymous · Answer

yea

whpalmer4 · Answer

can you give me the equation of a line which is the same as the x-axis?

whpalmer4 · Answer

it's just y = 0, right?

whpalmer4 · Answer

so when $g(t) = t^2-4$ crosses the $x$ axis, what is the value of $g(t)$?

anonymous · Answer

um yea

whpalmer4 · Answer

$$g(t) = 0$$when it crosses the $x$ axis, right?

anonymous · Answer

yea

whpalmer4 · Answer

So, we just need to solve $$g(t) = t^2-4 = 0$$for the value(s) of $t$ that make that true.

whpalmer4 · Answer

Any idea how to proceed with that?

anonymous · Answer

g(t)=t^2=4

whpalmer4 · Answer

No, we need to solve $$g(t) = 0$$which means $$t^2-4=0$$because $$g(t) = t^2-4$$

anonymous · Answer

so is it -4?

whpalmer4 · Answer

Is what -4?

anonymous · Answer

oh nvm

whpalmer4 · Answer

The sum of -2 and -2?  Yes, the sum of -2 and -2 is -4 :-)

anonymous · Answer

yea

whpalmer4 · Answer

we're trying to find values of $t$ that make $$t^2 -4 =0$$

whpalmer4 · Answer

what if we were just trying to solve $$t -4 = 0$$what would you do?

anonymous · Answer

plus 4 on both sides

anonymous · Answer

which gets you t^2=4 and then square root both sides and you get t=2

whpalmer4 · Answer

that's an answer.  are there any others?

anonymous · Answer

-2

whpalmer4 · Answer

very good.  so let's check our work:

$$g(t) = t^2- 4$$$$g(2) = (2)^2 -4 = 4-4=0\checkmark$$$$g(-2) = (-2)^2-4 = 4 - 4 = 0\checkmark$$

anonymous · Answer

wait i have a qusetion, why does it have to equal to 0 agian?

whpalmer4 · Answer

we wanted the points where it crosses the x-axis.  These are called "roots" or "zeros" or "solutions".  The x-axis is equivalent to the line y = 0.  If you want to find the points at which two functions intersect, you set the y values equal to each other and solve for the x values that make that true.  

$$y = 0$$$$y = t^2-4$$$$0=t^2-4$$$$t=\pm2$$

whpalmer4 · Answer

Here's a graph.

anonymous · Answer

Ohh i see thank you, can i ask you some more questions tomorrow? my mom said i need to go to sleep

whpalmer4 · Answer

I'll almost certainly be on tomorrow at some point. You know how to flag me down...I'll have a look later if I'm not on when you do so.

anonymous · Answer

when will you be on?

whpalmer4 · Answer

Don't know.

whpalmer4 · Answer

But I have a lot of work to do, so I'll probably be doing plenty of procrastinating by goofing around on OpenStudy :-)