maths question - leaving cert algebra

ublinina

Can anybody help with this leaving cert maths question. its algebra part c .

The quadratic equation f(x) = ax^2+ bx +c is shown , where a>0 ( a diagram of a quadratic curve is shown with the curve touching the x axis at point p ).

The curve touches the x-axis at the point p.

Find the co-ordinates of the point p in terms of a and b .

Pherekydes

If the curve touches the x-axis then f(x)=0, so the quadratic=0. The question wants you to derive the formula



You can find out how to do this at:

http://www.1728.com/quadr2.htm

ublinina

the quadratic formula x = -b+-sqrt b^2 - 4ac gives you the answer in terms of a, b and c. - i need the x part ( soln )in terms of a and b not in terms of a , b and c .

wolf1728

ublinina
Need some help with that problem? I found this site because Slow Coach linked that graphic from my site. (Some website owners hate that - I don't. Heck I even got credit for it. Okay nuff said).

Anyway if you need help just post here or send an E-Mail.

M@lice

For a quadratic equation f(x)=ax^2 + bx + c that touches the x axis at one point only b^2-4ac=0 (it follows from the quadratic formula). Also you know that f(p)=0. Using those two pieces of information should be enough to solve the problem.

Adama

Hey,

It's quite simple but the question seems to be worded wrong. If the quad curve only touches the x-axis at 1 point (i.e. P) then c = 0, by definition of a quad curve. Then the point P is given by:

P=(x,y)

y = 0 ...as it is on the x-axis

x =>
If y=ax^2 + bx +c ...Remember f(x) is the same as y
Then

ax^2 + bx = 0 ... As y=0, and c=0

Rearranging this with algebra you will get,

x = -b/a

So the point P is at (-b/a , 0)

Hope this helps.

[EDIT] you know this answer seems right because it tells you a > 0, so this means the value -b/a can always to found! [EDIT]

M@lice

It's quite simple but the question seems to be worded wrong. If the quad curve only touches the x-axis at 1 point (i.e. P) then c = 0, by definition of a quad curve. Then the point P is given by:

I'm afraid that is untrue in the majority of cases. A quadratic curve f(x) = ax^2 + bx + c touching the x axis at one point only implies that b^2 - 4ac = 0. c = 0 only implies that x = 0 is a root of the equation.

JimG

If the quadratic curve touches the x-axis tnen there is only one root for the quation f(x) = 0 (or: the two roots are equal). For this to happen we must have b^2 - 4ac = 0. The root then is -b/2a. c = 0 is irrelevant.

Pherekydes

JimG wrote:

If the quadratic curve touches the x-axis tnen there is only one root for the quation f(x) = 0 (or: the two roots are equal). For this to happen we must have b^2 - 4ac = 0. The root then is -b/2a. c = 0 is irrelevant.

b^2 - 4ac = 0

b^2 = 4ac

b = 2 Sqrt (ac) (Not -b/2a as stated above.)

C is very relevant. If you draw the graph c gives you the vertical displacement of the graph. If the graph touches the x-axis at one point this implies c=0. Then you can use the quadratic equation formula to solve for x.

This gives x = (-b/a, 0) as Adama stated (twice, in fact)

JimG

When b^2 - 4ac = 0, the single root is -b/2a. You can see this by substituting 0 under the square root sign in the solution formula.
The value of c is NOT relevant here. Try graphing, for example, x^2 - 2x + 1,
x^2 - 4x + 4, or any perfect square. Each graph touches the x-axis but c is not 0.
c gives vertical displacement for linear graphs.

Pherekydes

JimG wrote:

When b^2 - 4ac = 0, the single root is -b/2a. You can see this by substituting 0 under the square root sign in the solution formula.
The value of c is NOT relevant here. Try graphing, for example, x^2 - 2x + 1,
x^2 - 4x + 4, or any perfect square. Each graph touches the x-axis but c is not 0.
c gives vertical displacement for linear graphs.

I'm a bit rusty at the quadratics. Answer finally came and Jim is right.

If the graph touches the x-axis, this implies that the turning point of the graph is at point P. The slope of the tangent at P is zero. So differentiate, let the result = 0 and rearrange, to wit:

y = ax^2 + bx +c

dy/dx = 2ax + b

2ax + b = 0

2ax = -b

x = -b/2a

JimG

Full solutions, and the marking schemes, for the 2004 LC Higher Maths can be found at
http://www.examinations.ie/archive/markingschemes/2004/LC003ALP1EV.pdf
You will need Acrobat reader.