Answer: (a) intercepts \((-5,0)\), \(\left(\frac23,0\right)\), \((0,10)\); (b) \(\displaystyle k=0\) or \(\displaystyle k\gt \frac{289}{12}\).
Answer: (a) intercepts \((-5,0)\), \(\left(\frac23,0\right)\), \((0,10)\); (b) \(\displaystyle k=0\) or \(\displaystyle k\gt \frac{289}{12}\).
(a) First find where the graph meets the axes.
For the \(y\)-intercept, put \(x=0\):
\(y=\left|-10\right|=10.\)
So the graph meets the \(y\)-axis at
\((0,10).\)
For the \(x\)-intercepts, solve
\(3x^2+13x-10=0.\)
Factorising gives
\((3x-2)(x+5)=0.\)
Hence
\(x=\frac23\quad\text{or}\quad x=-5.\)
So the \(x\)-intercepts are
\(\left(\frac23,0\right)\quad\text{and}\quad(-5,0).\)
The graph is the quadratic \(y=3x^2+13x-10\), with the part below the \(x\)-axis reflected above the \(x\)-axis.
(b) The stationary point of
\(3x^2+13x-10\)
occurs when
\(x=-\frac{13}{2\cdot3}=-\frac{13}{6}.\)
At this point,
\(3\left(-\frac{13}{6}\right)^2+13\left(-\frac{13}{6}\right)-10=-\frac{289}{12}.\)
So the reflected maximum of the middle part of \(y=\left|3x^2+13x-10\right|\) is
\(\frac{289}{12}.\)
If \(k=0\), the equation has exactly the two intercept roots.
If \(0\lt k\lt \frac{289}{12}\), the horizontal line \(y=k\) cuts the graph in four distinct points.
If \(k=\frac{289}{12}\), it cuts the graph in three distinct points, because it is tangent at the reflected maximum.
If \(k\gt \frac{289}{12}\), it cuts the graph in exactly two distinct points.
Therefore
\(k=0\quad\text{or}\quad k\gt \frac{289}{12}.\)
