A-Level Edexcel Maths Pure Transformations of Functions: Show that $x^2+6x+11$ can be wri

Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$ where p and q are integers to be found - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 1

Question 6

Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$ where p and q are integers to be found. In the space at the top of page 7, sketch the curve with equation $y... show full transcript

Worked Solution & Example Answer:Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$ where p and q are integers to be found - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 1

Step 1

Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$

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Answer

To rewrite the expression in the form $(x + p)^2 + q$ , we need to complete the square.

Start with the quadratic expression:

$x^2 + 6x + 11$
To complete the square for $x^2 + 6x$ , take half of the coefficient of $x$ (which is 6), divide it by 2 to get 3, and then square it to get 9.

$x^2 + 6x = (x + 3)^2 - 9$
Substitute back into the original expression:

$(x + 3)^2 - 9 + 11 = (x + 3)^2 + 2$

Thus, we find that $p = 3$ and $q = 2$ .

Step 2

In the space at the top of page 7, sketch the curve with equation $y = x^2 + 6x + 11$, showing clearly any intersections with the coordinate axes.

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Answer

The equation $y = x^2 + 6x + 11$ can be sketched as follows:

Identify the vertex based on the rewritten form $(x + 3)^2 + 2$ .
The vertex is at (-3, 2), indicating the lowest point on the graph.
Intersections with the coordinate axes:
- Y-axis: Set $x = 0$ :
  
  $y = 0^2 + 6(0) + 11 = 11$
  Thus, the curve intersects the Y-axis at (0, 11).
- X-axis: Set $y = 0$ :
  
  $0 = x^2 + 6x + 11$ The discriminant of this quadratic, $ext{D} = b^2 - 4ac = 6^2 - 4(1)(11) = 36 - 44 = -8$ , shows there are no real intersections with the x-axis.

The sketch is a U-shaped curve opening upwards, with a vertex at (-3, 2) and a point on the Y-axis at (0, 11).

Step 3

Find the value of the discriminant of $x^2 + 6x + 11$

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Answer

The discriminant of the quadratic equation $ax^2 + bx + c$ is given by the formula:

$D = b^2 - 4ac$

For our equation, we identify:

$a = 1$
$b = 6$
$c = 11$

Now we substitute these values into the formula:

$D = 6^2 - 4(1)(11)$ $D = 36 - 44$ $D = -8$

Thus, the value of the discriminant is -8.

Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$ where p and q are integers to be found - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 1

Worked Solution & Example Answer:Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$ where p and q are integers to be found - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 1

Show that $x^2+6x+11$ can be written as $(x + p)^2 + q$

In the space at the top of page 7, sketch the curve with equation $y = x^2 + 6x + 11$, showing clearly any intersections with the coordinate axes.

Find the value of the discriminant of $x^2 + 6x + 11$

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