Exercise 28 Page 446

To graph the parabola, we first need to identify the axis of symmetry and the vertex. In this form, the axis of symmetry of the parabola is the vertical line $x=h$ and the vertex is the point $(h,0).$ Now consider the given function. We can see that $a=6$ and $h=\text{-}2.$ Therefore, the vertex is $(\text{-}2,0),$ and the axis of symmetry is $x=\text{-}2.$ To graph the function we need to find two more points on the graph. Let's choose two $x\text{-}$values less than the $x\text{-}$coordinate of the vertex and make a table of values.

$x$	$6(x+2{)}^2$	$q(x)=6(x+2{)}^2$
$\text{-}3$	$6(\text{-}3+2{)}^2$	$6$
$\text{-}4$	$6(\text{-}4+2{)}^2$	$24$

Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.

Let's draw a smooth curve that connects the five points. We will also draw the parent function $f(x)=x^2.$

From the graph above, we can note the following.

• Both graphs open up.
• The graph of $q(x)=6(x+2{)}^2$ is narrower than the graph of $f(x)=x^2.$
• The axis of symmetry of the parent function is the $y\text{-}$axis. The axis of symmetry of the given function is $x=\text{-}2.$
• The vertex of the given function, $(\text{-}2,0),$ is to the left of the vertex of the parent function, $(0,0).$

From the graph and the observations above, we can conclude that the graph of $q$ is a vertical stretch by a factor of $6$ and a horizontal translation left $2$ units of the graph of $f.$