so we have the points of (0,-7),(7,-14),(-3,-19), let's plug those in the y = ax² + bx + c form, since we have three points, we'll plug each one once, thus a system of three variables, and then we'll solve it by substitution.

$\bf \begin{array}{cccllll} \stackrel{\textit{point (0,-7)}}{-7=a(0)^2+b(0)+c}& \stackrel{point (7,-14)}{-14=a(7)^2+b(7)+c}& \stackrel{point (-3,-19)}{-19=a(-3)^2+b(-3)+c}\\\\ -7=c&-14=49a+7b+c&-19=9a-3b+c \end{array}$

well, from the 1st equation, we know what "c" is already, so let's just plug that in the 2nd equation and solve for "b".

$\bf -14=49a+7b-7\implies -7=49a+7b\implies -7-49a=7b \\\\\\ \cfrac{-7-49a}{7}=b\implies \cfrac{-7}{7}-\cfrac{49a}{7}=b\implies -1-7a=b$

well, now let's plug that "b" into our 3rd equation and solve for "a".

$\bf -19=9a-3b-7\implies -12=9a-3b\implies -12=9a-3(-1-7a) \\\\\\ -12=9a+3+21a\implies -15=9a+21a\implies -15=30a \\\\\\ -\cfrac{15}{30}=a\implies \blacktriangleright -\cfrac{1}{2}=a \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{and since we know that}}{-1-7a=b}\implies -1-7\left( -\cfrac{1}{2} \right)=b\implies -1+\cfrac{7}{2}=b\implies \blacktriangleright \cfrac{5}{2}=b \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{2}x^2+\cfrac{5}{2}x-7~\hfill$

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3 years ago