The graph of the equation \(y = a|x-b| + c\) has two x-intercepts at \(x=-3\) and \(x=7\),and one y-intercept at \(y=5\). What is the product abc?
+26367 The graph of the equation \(y = a|x-b| + c\) has two x-intercepts at \(x=-3\) and \(x=7\),and one y-intercept at \(y=5\).
What is the product \(abc\) ?
\(\begin{array}{|lrclrcl|} \hline \mathbf{x-\text{intercepts}\quad y=0:} & 0 &=& a|x-b| + c \\ & \mathbf{|x-b|}&=& \mathbf{-\dfrac{c}{a}} \\\\ & x&=& -3 & x&=&7 \\ & |-3-b|&=& -\dfrac{c}{a} & |7-b|&=& -\dfrac{c}{a} \\ & |-(3+b)|&=& -\dfrac{c}{a} & \left(7-b\right)^2&=& \left(-\dfrac{c}{a}\right)^2 \quad(1) \\ & |3+b|&=& -\dfrac{c}{a} \\ & \left(3+b\right)^2&=& \left(-\dfrac{c}{a}\right)^2 \quad(2) \\\\ (1)=(2): & \left(3+b\right)^2&=& \left(7-b\right)^2 \\ & 9+6b+b^2 &=&49-14b+b^2 \\ & 9+6b &=&49-14b \\ &20b &=& 40 \\ & \mathbf{b} &=& \mathbf{2} \qquad |b|=2 \\\\ & |3+b|&=& -\dfrac{c}{a} \\ & |3+2|&=& -\dfrac{c}{a} \\ & 5 &=& -\dfrac{c}{a} \\ & 5a &=& -c \\ & \mathbf{ a } &=& \mathbf{-\dfrac{c}{5}} \\ \hline \mathbf{y-\text{intercepts}\quad x=0:} & y &=& a|0-b| + c \\ & y &=& a|b| + c \\ & \mathbf{y} &=& \mathbf{2a + c} \\\\ & y &=& 5 \\ & 5 &=& 2a+c \quad | \quad a =-\dfrac{c}{5} \\ & 5 &=& 2\left(-\dfrac{c}{5}\right)+c \\ & 5 &=& -\dfrac{2}{5} c+c \\ & 5 &=& \dfrac{3}{5} c \\ & \mathbf{ c } &=& \mathbf{ \dfrac{25}{3}} \\\\ & a &=& -\dfrac{c}{5} \\ & a &=& -\dfrac{\dfrac{25}{3}}{5} \\ & \mathbf{ a} &=& \mathbf{- \dfrac{5}{3}} \\ \hline \end{array}\)
\(\boxed{\mathbf{y = \left( - \dfrac{5}{3}\right)|x-2| + \dfrac{25}{3}}} \\\)
\(\begin{array}{|rcll|} \hline abc &=& \left( - \dfrac{5}{3}\right)\cdot 2 \cdot \dfrac{25}{3} \\ \mathbf{abc} &=& \mathbf{-\left( \dfrac{250}{9}\right)} \\ \hline \end{array}\)
