Question 629298
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Find the quadratic polynomial whose graph goes through the points (−1,5), (0,5), and (2,29).
f(x)= _x2+_ x+_ <<< that form

I got to the step 4a+2b+0=29
that was like my 6/7th step. I am getting stuck from there. Can i get any help please?
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<pre>
Notice that the y-coordinate is the same, '5', for both points (-1,5) and (0,5).


It means that if you consider another quadratic function, g(x) = f(x) - 5, instead of f(x),
this new quadratic function will have x-intercepts at  x= -1  and  x= 0.


In turn, it means that g(x) = ax*(x+1) with some real coefficient 'a',

so f(x) = ax(x+1) + 5.


Great (!)   So, to determine f(x), we only need to find single unknown coefficient 'a'.


For it, we use the condition that the parabola passes through the point (2,29).


It gives us this equation

    f(2) = 29,  or  a*2*(2+1) + 5 = 29,  or  2*3*a = 29 - 5,  or  6a = 24.


Thus we find  a = 24/6 = 4.


So, our quadratic function is  f(x) = 4x(x+1) + 5,  or  f(x) = 4x^2 + 4x + 5.
</pre>

At this point, the problem is solved completely, and we obtained the solution 
with minimal computations and maximal fun.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This trick works for many other similar problem, where 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;two points of the three given points have the same y-coordinate.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So, &nbsp;this method is worth to learn &nbsp;(&nbsp;!&nbsp;)