please help!!
Hello,
to #2:
The coordinates of the point must satisfy the equation of the function. That means if you plug in the coordinates the equation must be true:
$\displaystyle 5 = (2-1)^2 \cdot (2+a)~ \Longrightarrow ~ 5 = 1 \cdot (2+a) ~ \Longrightarrow ~ a = 3$
The function becomes: $\displaystyle y = f(x) = (x-1)^2 \cdot (x+3)$.
This is a cubic function with the domain $\displaystyle \mathbb{R}$. The y-values are running from $\displaystyle -\infty$ to $\displaystyle +\infty$.
The function has a zero at x = -3 and a second zero at x = 1 where the graph touches the x-axis. Therefore:
y < 0 for x < -3 and y > 0 for x > -3
To draw the function $\displaystyle g(x) = |f(x)|$ reflect the parts of the graph which are below the x-axis over / at(?) the x-axis:
$\displaystyle g(x) = \left\{ \begin{array}{c}-f(x), x < -3 \\f(x), x \geq -3\end{array} \right.$
The graph of g is sketched in red.