Question 178109
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If a quadratic function *[tex \Large f(x) = ax^2 + bx + c] has a graph containing a point *[tex \Large (x_1,y_1)] then *[tex \Large ax_1^2 + bx_1 + c = y_1].


So, since *[tex \Large f(x) = ax^2 + bx + c] contains the point *[tex \Large (0, 3)] we know that *[tex \Large a(0)^2 + b(0) + c = 3] which is to say *[tex \Large c = 3].


Similarly, since *[tex \Large f(x) = ax^2 + bx + c] contains the point *[tex \Large (1, -4)] we know that *[tex \Large a(1)^2 + b(1) + c = -4].  But since we already know that *[tex \Large c = 3], we can say *[tex \Large a + b + 3 = -4] which is to say *[tex \Large a + b = -7].


Continuing the process, the third point, *[tex \Large (2, -9)], will result in: *[tex \Large a(2)^2 + b(2) + 3 = -9] which is to say *[tex \Large 4a + 2b = -12].


Now:


*[tex \Large \text {         } \math a + b = -7] and


*[tex \Large \text {         } \math 4a + 2b = -12]


form a system of two linear equations in two variables.  Given that the coefficents on both variables in the first equation is 1, this system lends itself well to solution by substitution.


*[tex \Large a + b = -7 \text { } \Rightarrow \text{ }\math a = -b - 7]


Substituting this expression for <i>a</i> into the second equation yields:


*[tex \Large 4(-b-7) + 2b = -12 \text { } \Rightarrow \text{ }\math  -4b -28 + 2b = -12 \text { } \Rightarrow \text{ }\math -2b = 16 \text { } \Rightarrow \text{ }\math b = -8]


Now that we know *[tex \Large b = -8] we can substitute this value into *[tex \Large a = -b - 7 \text { } \Rightarrow \text{ }\math a = -(-8) - 7 = 1]


Now that we have values for <i>a</i>, <i>b</i>, and <i>c</i>, we can write the quadratic function specifically:


*[tex \Large f(x) = ax^2 + bx + c = x^2 - 8x + 3]    


Do the other problem the same way.
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