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# completing the square

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AQA GCSE physics P1 unofficial mark scheme 05-05-2016
1. 1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.

2. Hence show that x2 + 2x + 5 is always positive.
2. presuming you want help with this ??
3. (Original post by LoveLifeHate)
presuming you want help with this ??
4. x^2 + 2x + 5 = (x + 1)^2 + 4

y = (x + 1)^2 + 4

dy/dx = 2x + 2

at turning point, 0 = 2x + 2
x=-1

at x=-1, y=4

Because of the shape of the graph, the turning point is always a minimum point, so the lowest point is greater than 0.
5. I always remember HALF, SQUARE, COPY, DONE when completing the square.

for x^2+bx+c, half b, square b so that you have (x+b/2)^2 - b^2 +c. You should have been told how to interpret a curve in this form.
6. (Original post by WarriorInAWig)
I always remember HALF, SQUARE, COPY, DONE when completing the square.

for x^2+bx+c, half b, square b so that you have (x+b/2)^2 - b^2 +c. You should have been told how to interpret a curve in this form.
7. (Original post by A.J10)
x^2 + 2x + 5 = (x + 1)^2 + 4

y = (x + 1)^2 + 4

dy/dx = 2x + 2

at turning point, 0 = 2x + 2
x=-1

at x=-1, y=4

Because of the shape of the graph, the turning point is always a minimum point, so the lowest point is greater than 0.
sorry but how is the minimum point greater than o when x=-1?

also, q1 the equation is y = x2 + 10x + 19.
8. (Original post by non)
sorry but how is the minimum point greater than o when x=-1?
When x=-1, f(x)=4. Since (-1, 4) is the minimum/vertex, there does not exist a point on the curve that has a y-coordinate less than 4. Thus, f(x)≥4, and by extension f(x)=x^2+2x+5>0.

At a glance, I have no clue how question 2 follows from question 1...
9. (Original post by A.J10)
x^2 + 2x + 5 = (x + 1)^2 + 4

y = (x + 1)^2 + 4

dy/dx = 2x + 2

at turning point, 0 = 2x + 2
x=-1

at x=-1, y=4

Because of the shape of the graph, the turning point is always a minimum point, so the lowest point is greater than 0.
You could have stopped at the first line.

10. (Original post by non)
1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.
Given the function

,

completing the square on the RHS tells us that

Spoiler:
Show

Essentially, you have placed the quadratic into vertex form.

It follows that the x-coordinate of the parabola y=ax+by+c's vertex is always .

You should know that the line of symmetry for a parabola is a vertical line (vertical lines have the equation x=?) and that it crosses the vertex (so we know one x value of the equation x=?).

As such, the equation for said vertical line can be reasoned out.
11. (Original post by non)
1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.

2. Hence show that x2 + 2x + 5 is always positive.
1.

is a quadratic so symmetry occurs at its unique turning point.

Therefore turning point at is the line of symmetry of .

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