anonymous
  • anonymous
How do I find the function of this graph?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
let's see: \[f(x) = x ^{2} - 2x - 3\]
tatianagomezb
  • tatianagomezb
Find the roots and then replace from this formula f (x) = a (r -x) *(r-x)

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anonymous
  • anonymous
@Cardinal_Carlo @tatianagomezb thank you!
tatianagomezb
  • tatianagomezb
You're welcome! I hope it helped!
whpalmer4
  • whpalmer4
@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:
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tatianagomezb
  • tatianagomezb
@whpalmer4 I like keeping it simple!
whpalmer4
  • whpalmer4
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
  • tatianagomezb
But it can't be -1, because we can all agree the parabola is pointing upwards.
anonymous
  • anonymous
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
  • whpalmer4
@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
  • tatianagomezb
Oh, yes. You're right about that. @whpalmer4 my teacher mentioned I tend to override important stuff. @cardinal_carlo you're right too.

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