completing the square

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.

Reply 1

LoveLifeHate

presuming you want help with this ??

Reply 2

non

OP

15

Original post by LoveLifeHate

presuming you want help with this ??

yes please

Reply 3

A.J10

14

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.

Reply 4

WarriorInAWig

15

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.

Reply 5

A.J10

14

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.

Additionally you can expand it out in your head to make sure it matches the original.

Reply 6

non

OP

15

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.

(edited 11 years ago)

Reply 7

aznkid66

1

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...

Reply 8

Lord of the Flies

19

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.

(x+1)^2+4>0

Reply 9

aznkid66

1

Original post by non

1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.

Given the function

y = ax^2 + bx + c

,

completing the square on the RHS tells us that

Spoiler

y=a\left(x-\left(-\dfrac{b}{2a}\right)\right)^2 + \dfrac{-b^2+4ac}{4a}

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

\dfrac{-b}{2a}

.

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.