f(x) = x² - 8x + 19 (a) Express f(x) in the form (x + a)² + b, where a and b are constants - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 1

To sketch the graph of C, we need to identify the key points:

1. The curve $y = f(x)$ is a parabola that opens upwards with its vertex at the point Q, which we've identified at (4, 3).

2. To find the y-intercept P, we set x = 0:

   $f(0) = (0 - 4)² + 3 = 16 + 3 = 19$ So the coordinates of P are (0, 19).

3. The graph should clearly show:

   • Point P at (0, 19): the y-intercept.
   • Point Q at (4, 3): the minimum point.

4. The sketch should resemble a U-shaped curve, with these two points highlighted.

To find the distance PQ between the two points P(0, 19) and Q(4, 3), we can use the distance formula:

$d = ext{PQ} = ext{sqrt}((x_2 - x_1)² + (y_2 - y_1)²)$

Substituting the coordinates:

• Let (x₁, y₁) = (0, 19) and (x₂, y₂) = (4, 3).

1. Calculate the differences:

   • $x_2 - x_1 = 4 - 0 = 4$
   • $y_2 - y_1 = 3 - 19 = -16$

2. Now substitute into the distance formula:

   $d = ext{sqrt}((4)² + (-16)²)$

   $d = ext{sqrt}(16 + 256) = ext{sqrt}(272)$

3. Simplifying this:

   $ext{sqrt}(272) = ext{sqrt}(16 imes 17) = 4 ext{sqrt}(17)$ Thus, the distance PQ = $4 ext{sqrt}(17)$.