$x^2 - 4x + 5 = (x - a)^2 + b$ Find the value of $a$ and of $b$. Answer a = ... b = ... The curve $y = x^2 - 4x + 5$ is drawn. Write down the equation of the line of symmetry of the curve. Answer ... Sketch the curve $y = x^2 - 4x + 5$ on the axes below. Indicate clearly the coordinates of the points where the graph crosses the axes and the minimum point on the curve. Answer

1. Expand the completed-square form and match the coefficient of $x$.

$(x-a)^2+b=x^2-2ax+a^2+b$

Compare with
$x^2-4x+5.$

So
$-2a=-4$
$a=2.$

2. Substitute $a=2$ and match the constant term to find $b$.

Using $a=2$,
$(x-a)^2+b=(x-2)^2+b=x^2-4x+4+b.$

Compare the constant term with $x^2-4x+5$:
$4+b=5$
$b=1.$

1. Identify the line of symmetry from the completed-square form.

From part (a),
$x^2-4x+5=(x-2)^2+1.$
For $y=(x-a)^2+b$, the line of symmetry is $x=a$.

Hence the line of symmetry is
$x=2.$

1. Use the completed-square form to identify the minimum point and shape of the curve.

From part (a),

$y=x^2-4x+5=(x-2)^2+1.$

Since the coefficient of $x^2$ is positive, the parabola opens upwards.

The minimum point is

$(2,1).$

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

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

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

So the graph crosses the y-axis at

$(0,5).$

3. Check whether the graph crosses the x-axis.

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

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

Using the completed-square form,

$(x-2)^2+1=0.$

But $(x-2)^2 \geqslant 0$, so $(x-2)^2+1>0$ for all real $x$.

Therefore, the graph does not cross the x-axis.

4. Sketch and label the required features clearly.

Draw a smooth U-shaped curve opening upwards, symmetric about $x=2$, with minimum point $(2,1)$. The curve passes through $(0,5)$ on the y-axis and has no x-axis crossing.