Exercise 55 Page 146

We want to draw the graph of the given quadratic function. To do so, we will rewrite it in vertex form, $g(x)=a(x-h{)}^2+k,$ where $a,$ $h,$ and $k$ are either positive or negative numbers. To draw the graph, we will follow four steps.

1. Identify the constants $a,$ $h,$ and $k.$
2. Plot the vertex $(h,k)$ and draw the axis of symmetry $x=h.$
3. Plot any point on the curve and its reflection across the axis of symmetry. We can also find the $x\text{-}$intercepts.
4. Sketch the curve.

Let's get started.

Step $1$

We will first identify the constants $a,$ $h,$ and $k.$ Recall that if $a<0,$ the parabola will open downwards. Conversely, if $a>0,$ the parabola will open upwards. We can see that $a=1,$ $h=2,$ and $k=\text{-}4.$ Since $a$ is greater than $0,$ the parabola will open downwards.

Step $2$

Let's now plot the vertex $(h,k)$ and draw the axis of symmetry $x=h.$ Since we already know the values of $h$ and $k,$ we know that the vertex is $(2,\text{-}4).$ Therefore, the axis of symmetry is the vertical line $x=2.$

Step $3$

We will now plot a point on the curve by choosing an $x\text{-}$value and calculating its corresponding $y\text{-}$value. We can also find the $x\text{-}$intercepts and plot them. Since we want to find the $x\text{-}$intercepts, we will find them now and plot them. To do this, we will solve the equation $g(x)=0.$ Now we can calculate the first $x\text{-}$intercept using the positive sign and the second one using the negative sign.

$x=2\pm 2$
$x=2+2$	$x=2-2$
$x=4$	$x=0$

Therefore, the $x\text{-}$intercepts are $x=4$ and $x=0.$

Note that the point $(0,0),$ which is an $x\text{-}$intercept, is also the $y\text{-}$intercept.

Step $4$

Finally, we will sketch the parabola which passes through the three points. Remember not to use a straightedge for this!