Answer:
\((5,-24)\) and \((3,-12)\)
Step-by-step explanation:
we have
\(y=-x^{2} +2x-9\) -----> equation A
\(y=-6x+6\) -----> equation B
equate the equation A and equation B
\(-x^{2} +2x-9=-6x+6\)
\(-x^{2} +2x-9+6x-6=0\)
\(-x^{2} +8x-15=0\)
The formula to solve a quadratic equation of the form \(ax^{2} +bx+c=0\) is equal to
\(x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}\)
in this problem we have
\(-x^{2} +8x-15=0\)
so
\(a=-1\\b=8\\c=-15\)
substitute in the formula
\(x=\frac{-8(+/-)\sqrt{8^{2}-4(-1)(-15)}} {2(-1)}\)
\(x=\frac{-8(+/-)\sqrt{64-60}} {-2}\)
\(x=\frac{-8(+/-)2}{-2}\)
\(x=\frac{-8+2}{-2}=3\)
\(x=\frac{-8-2}{-2}=5\)
Find the values of y
For \(x=3\)
\(y=-6x+6\) ------> \(y=-6(3)+6=-12\)
For \(x=5\)
\(y=-6x+6\) ------> \(y=-6(5)+6=-24\)
The solutions are the points \((3,-12)\) and \((5,-24)\)
