Hi, I cannot seem to wrap my head around this question! Could someone please show the steps to get the answer to this question? Thank you!!
The answers are:
f(x)=−2(x−1)(x+5)
g(x) = −50(x − 1)( x + 1/5)
Given that the conditions stated are identical and you are asked to find two quadratic functions for which this is true, we can start with one general quadratic. We know that \(x-1\) is a factor, due to the fact that \(f\left(1\right)=g\left(1\right)=0\). We don't yet know the other factor, or the dilation factor, thus:
&=A\left(x-1\right)\left(x-k\right)\\&=A\left(x^2-x-kx+1\right)\\&=A\left(x^2-\left(1+k\right)x+1\right)\end{align*})
The \(x\)-value of the turning point of any quadratic in expanded form is always \(-\frac{b}{2a}\) where \(b\) is the coefficient of \(x\) and \(a\) is the coefficient of \(x^2\) and in this particular case, the \(y\)-value of the turning points of \(f\left(x\right)\) and \(g\left(x\right)\) is 18:
}{2}\\&=\frac{1+k}{2}\\\therefore 18&=A\left(\frac{1+k}{2}-1\right)\left(\frac{1+k}{2}-k\right)\\&=A\left(\frac{1}{2}\left(1+k-2\right)\right)\left(\frac{1}{2}\left(1+k-2k\right)\right)\\&=\frac{A}{4}\left(k-1\right)\left(1-k\right)\\&=-\frac{A}{4}\left(k-1\right)^2\end{align*})
We also know the \(y\)-intercept:
\\&=A\left(-1\right)\left(-k\right)\\&=Ak\end{align*})
We now have a pair of simultaneous equations that can be solved:
^2\ \left(1\right)\\72&=-A\left(k-1\right)^2\ \left(1a\right)\\10&=Ak\ \left(2\right)\\\frac{10}{k}&=A\ \left(2a\right)\\\text{Substitute }\left(2a\right)&\rightarrow\left(1a\right):\\72&=-\frac{10}{k}\left(k-1\right)^2\\-72k&=10\left(k^2-2k+1\right)\\&=10k^2-20k+10\\0&=10k^2+52k+10\\0&=5k^2+26k+5\\&=\left(5k^2+25k\right)+\left(k+5\right)\\&=5k\left(k+5\right)+\left(k+5\right)\\&=\left(5k+1\right)\left(k+5\right)\\k_1&=-\frac{1}{5}\\k_2&=-5\\\text{Substitute }k_1&\rightarrow\left(2a\right):\\A_1&=\frac{10}{-\frac{1}{5}}\\&=-50\\\text{Substitute }k_2&\rightarrow\left(2a\right):\\A_2&=\frac{10}{-5}\\&=-2\end{align*})
One of these solution pairs belongs to \(f\left(x\right)\) and the other to \(g\left(x\right)\), but the question doesn't specify which is which, so we could have:
&=-50\left(x-1\right)\left(x+\frac{1}{5}\right)\\g\left(x\right)&=-2\left(x-1\right)\left(x+5\right)\\&\text{or}\\g\left(x\right)&=-50\left(x-1\right)\left(x+\frac{1}{5}\right)\\f\left(x\right)&=-2\left(x-1\right)\left(x+5\right)\end{align*})
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