4. (a) 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
4. (a) Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found.
(b) In the space at the top of page 7, sketch the cur... show full transcript
Worked Solution & Example Answer:4. (a) 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 express the quadratic x2+6x+11 in the form (x+p)2+q, we start by completing the square:
Identify the coefficient of x, which is 6. Half of this value is 3, so we can write:
x2+6x=(x+3)2−9
Substituting this back into the original expression yields:
x2+6x+11=(x+3)2−9+11
Simplifying further gives:
x2+6x+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
To sketch the curve y=x2+6x+11, we need to determine intersections with the axes:
Finding the y-intercept: Set x=0:
y=02+6(0)+11=11
Therefore, the curve intersects the y-axis at (0,11).
Finding the x-intercepts: To find x-intercepts, set y=0:
0=x2+6x+11
To solve for x, calculate the discriminant:
D=b2−4ac=62−4(1)(11)=36−44=−8
Since the discriminant is negative, there are no real x-intercepts. Thus, the curve does not cross the x-axis.
Sketching the curve: The curve is a U-shaped parabola opening upwards with a minimum point at (−3,2), as we derived in part (a). The vertex is above the x-axis, showing its location clearly.
Step 3
Find the value of the discriminant of $x^2 + 6x + 11$
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Answer
To find the discriminant of the quadratic x2+6x+11, we will use the formula:
