Express $x^2 + 5x + 4$ in the form $(x + p)^2 + q$. Answer (a) ... Sketch the graph of $y = x^2 + 5x + 4$. Indicate clearly the values where the graph crosses the x- and y- axes. Answer (b) Write down the coordinates of the minimum point of the graph of $y = x^2 + 5x + 4$. Answer (c) ( , )

1. Form the square using half the coefficient of $x$.

Half of $5$ is $\dfrac{5}{2}$, so
$x^2+5x+4=\left(x+\dfrac{5}{2}\right)^2+q.$

2. Adjust the constant term to keep the expression equivalent.

Since
$\left(x+\dfrac{5}{2}\right)^2=x^2+5x+\dfrac{25}{4},$
$x^2+5x+4=\left(x+\dfrac{5}{2}\right)^2-\dfrac{25}{4}+4$
$=\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}.$

1. Find the x-intercepts by setting $y=0$.

At the x-axis, $y=0$.

$x^2+5x+4=0$

$(x+1)(x+4)=0$

So $x=-1$ or $x=-4$.

The graph crosses the x-axis at $(-1,0)$ and $(-4,0)$.

2. Find the y-intercept by setting $x=0$, then sketch the parabola with the correct shape and labelled intercepts.

At the y-axis, $x=0$.

$y=0^2+5(0)+4=4$

The graph crosses the y-axis at $(0,4)$.

Since the coefficient of $x^2$ is positive, the graph is a U-shaped parabola opening upwards. Sketch it passing through $(-4,0)$, $(-1,0)$ and $(0,4)$.

1. Read the minimum point from the completed-square form.

From part (a),
$y=x^2+5x+4=\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}.$
The square term is smallest when
$x+\dfrac{5}{2}=0 \quad \Rightarrow \quad x=-\dfrac{5}{2}.$
Then
$y=-\dfrac{9}{4}.$