Express $y = -(x + 2)(x - 5)$ in the form of $y = -(x + p)^2 + k$. State the value of $p$ and the value of $k$. Answer: p = …. k = …. Sketch the graph of $y = -(x + 2)(x - 5)$ in the given axes below. Indicate any intercepts on the axes. Find the coordinates of the turning point of the graph $y = -(x + 2)(x - 5)$. Answer: (… , …)

1. Expand the product and simplify the quadratic expression.

$y=-(x+2)(x-5)$
$(x+2)(x-5)=x^2-3x-10$
$y=-(x^2-3x-10)=-x^2+3x+10$

2. Factor out the negative coefficient and complete the square inside the bracket.

$y=-(x^2-3x-10)$
Complete the square for $x^2-3x$:
$x^2-3x=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}$
So
$x^2-3x-10=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}-10$
$=\left(x-\frac{3}{2}\right)^2-\frac{49}{4}$

3. Rewrite in the required form and identify $p$ and $k$.

$y=-\left[\left(x-\frac{3}{2}\right)^2-\frac{49}{4}\right]$
$y=-\left(x-\frac{3}{2}\right)^2+\frac{49}{4}$
Comparing with $y=-(x+p)^2+k$,
$x+p=x-\frac{3}{2}$
$p=-\frac{3}{2},\qquad k=\frac{49}{4}$

1. Find the x-intercepts from the factorised quadratic.

For the x-intercepts, set $y=0$:

$-(x+2)(x-5)=0$

So

$x+2=0 \quad \text{or} \quad x-5=0$

$x=-2 \quad \text{or} \quad x=5$

The x-intercepts are $(-2,0)$ and $(5,0)$.

2. Find the y-intercept by substituting $x=0$.

For the y-intercept, put $x=0$:

$y=-(0+2)(0-5)$

$y=-(2)(-5)=10$

The y-intercept is $(0,10)$.

3. Sketch the correct parabola and label the intercepts.

Since

$y=-(x+2)(x-5)$

has a negative leading coefficient, the parabola opens downwards.

Sketch a smooth downward-opening parabola passing through $(-2,0)$, $(5,0)$ and $(0,10)$. The maximum point is at

$\left(\frac{3}{2},\frac{49}{4}\right)$

from part (a), which helps position the top of the curve.

1. Identify the turning point from the completed-square form.

From part (a),
$y=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{49}{4}.$
Hence the turning point is
$\left(\dfrac{3}{2},\dfrac{49}{4}\right).$