Express $x^2-4x+8$ in the form $(x+p)^2+q$. Answer ... Hence explain why there is no solution for $x^2-4x+8=0$. Answer ... ... Sketch the graph of $y=(3-x)(x+5)$ on the axes below. Indicate clearly the values where the graph crosses the axes and the coordinates of the turning point. Answer

1. Complete the square for the quadratic expression.

$x^2-4x+8=(x-2)^2-4+8=(x-2)^2+4$

1. Use the completed-square form to show the expression is always positive.

From part (a)(i),
$x^2-4x+8=(x-2)^2+4$.

Since $(x-2)^2\geq 0$ for all real $x$,
$(x-2)^2+4\geq 4>0.$
Hence $x^2-4x+8$ can never be $0$, so the equation has no solution.

1. Find where the graph crosses the $x$-axis.

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

$(3-x)(x+5)=0$

So

$3-x=0 \quad \text{or} \quad x+5=0$

$x=3 \quad \text{or} \quad x=-5$

The graph crosses the $x$-axis at $(3,0)$ and $(-5,0)$.

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

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

$y=(3-0)(0+5)$

$y=15$

The graph crosses the $y$-axis at $(0,15)$.

3. Find the turning point and determine the shape of the parabola.

The $x$-coordinate of the turning point is halfway between the two roots $-5$ and $3$.

$x=\frac{-5+3}{2}=-1$

Substitute $x=-1$:

$y=(3-(-1))(-1+5)$

$y=4\times 4=16$

So the turning point is $(-1,16)$.

Also,

$y=(3-x)(x+5)=-x^2-2x+15$

Since the coefficient of $x^2$ is negative, the parabola opens downwards.