Show that $y = 5 - x^2 - 4x$ has a maximum point $(-2, 9)$. Answer Sketch the graph of $y = 5 - x^2 - 4x$ on the axes below. Indicate clearly the values where the graph crosses the axes and the maximum point on the graph. Hence, explain why the equation $x^2 + 4x + 5 = 0$ does not have any solutions. Answer

1. Rewrite the quadratic by completing the square.

$y=5-x^2-4x$
$y=5-(x^2+4x)$
$y=5-\left[(x+2)^2-4\right]$
$y=9-(x+2)^2$

2. Use the completed-square form to show the maximum point.

Since $(x+2)^2\geq 0$, $-(x+2)^2\leq 0$.

Therefore,
$y=9-(x+2)^2\leq 9.$

The maximum value of $y$ is $9$, and it occurs when
$x+2=0,$
$x=-2.$

1. State the shape and maximum point of the quadratic graph.

From part (a),
$y=5-x^2-4x=9-(x+2)^2.$
Since the coefficient of $x^2$ is negative, the parabola opens downwards. Its maximum point is $(-2,9)$.

2. Find the intercept where the graph crosses the $y$-axis.

At the $y$-axis, $x=0$.
$y=5-0^2-4(0)=5.$
So the graph crosses the $y$-axis at $(0,5)$.

3. Find the intercepts where the graph crosses the $x$-axis and complete the sketch.

At the $x$-axis, $y=0$.
$5-x^2-4x=0$
$x^2+4x-5=0$
$(x+5)(x-1)=0$
So $x=-5$ or $x=1$.

The graph crosses the $x$-axis at $(-5,0)$ and $(1,0)$. Sketch a smooth downward-opening parabola through $(-5,0)$, $(0,5)$ and $(1,0)$, with maximum point $(-2,9)$.

1. Relate the given equation to the earlier graph.

Given
$x^2+4x+5=0,$
we get
$x^2+4x=-5.$
Hence
$5-x^2-4x=5-(x^2+4x)=5-(-5)=10.$

2. Use the maximum value of the graph to justify that there is no solution.

From part (a), the maximum value of
$y=5-x^2-4x$
is $9$.

So the graph cannot have $y=10$. Therefore there is no value of $x$ for which
$5-x^2-4x=10,$
and hence $x^2+4x+5=0$ has no solutions.