Quadratic and the parabola (1 Viewer)

[AnAn]

1) the parabola y=ax^2 + bx + c has its vertex at (2, 1) and passes through the point (0 ,0). find a, b and c.

2) x^2 + 10x - 17 = P( x - 2)^2 + Q(2x - 3) + R
i have found P=1, Q=7 and R=2
Now how do i express x^2 + 10x - 17 in the form P( x - 2)^2 + Q(2x - 3) + R ??

 

[untamedanimal]

1) Y = ax^2 + bx + c
when x = 0, y = 0 so c = 0

y = ax^2 + bx (divide by a)
Y/a = x^2 + bx/a (complete square)
Y/a + (b/2a)2 = x^2 + bx/a + (b/2a)^2
Y/a + (b/2a)2 = (x + b/2a)^2
1/a (Y + b^2/4a) = (x + b/2a)^2

for the vertices
1. B^2/4a = -1
2. B/2a = -2 (simultaneous equations)

B = -4a
(sub into 1)
64a2/4a = 1

16a = 1
a = 1/16
b = -1/4

I might be getting it wrong because im half asleep right now

 

[Rahul]

1)
draw the graph from the info you have. vertext at (2,1) and passes through the origin.

sub in both points into the equation: y=ax^2 + bx + c

you should end up with 1 = 4a + 2b + c and c = 0.

then, the trick of the question is to see that the graph is also going to pass through the point (4,0) due to symmetry. you need to use the graph from before.

then you sub in that point into the equation of the parabola. you should end up with 0 = 16a + 4b + c.

now you have 3 equations:
1 = 4a + 2b + c (1)
c = 0 (2)
0 = 16b + 4b + c (3)

subbing (2) into (1) and (3), you get:
1 = 4a + 2b (1A)
0 = 16b + 4b (3A)

solving them simultaneously, you should get: b = 1, a = -1/4 and previously, c = 0.

edit: mine might be wrong? dunno cbf checking...

 

[CM_Tutor]

You don't need to recognise the symmetry. The point (2, 1) is the vertex, and so at x = 2, dy/dx = 0.

Now, dy/dx = 2ax + b, and so it follows that 4a + b = 0

Now, put b = -4a, c = 0 and the point (2, 1) into the equation of the parabola, and you get: 1 = a(2)² -4a(2) + 0

So, a = -1 / 4, b = 1, c = 0

 

[Rahul]

yeh, and there is the smart way of doing it....
nice solution CM_Tutor.

 

[nike33]

isnt that the standard way of doing it?

 

[Rahul]

probably standard for the likes of you