How do I find the function of this graph?

Question

anonymous · Answer

attachment

anonymous · Answer

let's see:
$$f(x) = x ^{2} - 2x - 3$$

tatianagomezb · Answer

Find the roots and then replace from this formula f (x) = a (r -x) *(r-x)

anonymous · Answer

@Cardinal_Carlo @tatianagomezb thank you!

tatianagomezb · Answer

You're welcome!  I hope it helped!

whpalmer4 · Answer

@tatianagomezb's approach works very well when you can see the zero-crossings like this.  I'll show you another method that works even if you can't see them, and can be extended to higher order polynomials.

It's a parabola, so it can be written $$y = ax^2+bx+c$$We just need to find values of $a,b,c$ that make the parabola fit the graph.  Pick the vertex and plug its $x,y$ values into the prototype equation:

vertex is at $(1,-4)$:
$$-4 = a(1)^2 + b(1) + c$$$$-4 = a+b+c$$

Now pick two more points and do the same.
x-axis crossing at $(-1,0)$:
$$0=a(-1)^2 + b(-1) + c$$$$0 = a-b+c$$

x-axis crossing at $(3,0)$:
$$0=a(3)^2 + b(3) + c$$$$0 = 9a + 3b + c$$

Now we have 3 equations in 3 unknowns:
$$−4=a+b+c$$$$0=a−b+c$$$$0=9a+3b+c$$

If we add the first two together, we get$$-4+0 = a+a+b-b + c + c$$$$-4 = 2a+2c$$$$-2=a+c$$
Now we plug that into the first equation:
$$-4 = a + b + c$$$$-4 = (a+c) + b$$$$-4 = -2 + b$$$$b=-2$$

Now our remaining equations are
$$−4=a-2+c$$$$0=a−(-2)+c$$$$0=9a+3(-2)+c$$or after simplification
$$-2 = a+c$$$$6 = 9a+c$$
If we subtract those two equations, we get
$$-2-6 = a-9a + c-c$$$$-8 = -8a$$$$a = 1$$

and that leaves us with $$-2 = (1)+c$$$$c=-3$$

So our equation is $$y = 1x^2 - 2x -3$$and if you plug in some values you'll see that it matches the graph.  I've plotted it here:

tatianagomezb · Answer

@whpalmer4 I like keeping it simple!

whpalmer4 · Answer

Ah, but what if the parabola had the same vertex, but $a=-1$?  Not so simple any longer without any x-axis intercepts, right? :-)

tatianagomezb · Answer

But it can't be -1, because we can all agree the parabola is pointing upwards.

anonymous · Answer

Both of you are right. @whpalmer4 approach is, in fact, solid. But since we were granted two intercepts and a vertex, it's highly permissible to skip that lengthy yet inarguably thorough process. Well, as long as @vvercetti gets the right answer, it's all good.

whpalmer4 · Answer

@tatianagomezb my point is that if you had to do this with a parabola that did not cross the x-axis, your method would not work, but mine will.

tatianagomezb · Answer

Oh, yes. You're right about that. @whpalmer4 my teacher mentioned I tend to override important stuff.  @cardinal_carlo you're right too.