A-levels maths question (Differentiation) - Help?

Question

Hey, this isn't homework I'm studying for my A-level exam in January and don't understand this type of question and can't ask my teacher because we're off school. I would use simultaneous equations but there's 3 unknown variables.

The curve with equation y=ax^2 + bx + c passes through the point (1,2). The gradient of the curve is zero (0) at the point (2,1). Find the values of a,b,c.

I'd appreciate it if you even told me HOW to go about it without necessarily providing an answer.

Thanx Johnson, you've got best answer for answering first. Thanks anyone else whose answered.

Answer 1

I wish we could draw in this space. Try to visualize a general curve of the form y=ax^2 + bx + c ... what do you see in your mind? (pause and think before reading on)
.
.
.
All right, you should be imagining a parabola. The sign of "a" indicates whether it is curved up or down, with a positive "a" giving an upward curve and "b" giving a downward curve. We know that the "tip" of the parabola is the point at which the gradient - the slope - is zero. So the "tip" of this parabola must be at the point (2,1). And it must curve upwards, because the parabola passes through (1,2).

So without writing down any algebra or calculus, we can picture the solution - the parabola we want - pretty well. This means that when we get our answer, we can easily check it: a, b, and c should specify a parabola that has its tip at (2,1) and curves up to meet (1,2),

You could try to solve it now, then come back and check.

If you get stuck, I will first give a recipe for solving this, then I'll solve it. Try following the recipe to solve it before you look at the solution.

All right, now, let's be more formal: we'll find the gradient and set it equal to zero at the specified point:


Differentiate a x^2 + b x + c = y. The result is the gradient, dy/dx.

Set dy/dx=0. This will give you one equation relating a, x, and b.

Plug in the given value (2) for x at the point (2,1). Now you have an equation relating a and b.

We also know that y= 2 when x = 1. You can plug that in to the original equation and find an equation relating a,b, and c. We now have two equations for three unknowns. Not enough!


What are we missing? (think for a moment)


We also know the point (2,1) is on the curve! So we know
that 2^2 a + 2 b + c = 1. That's our third equation.

Solve the three equations in three unknowns to recover a, b, and c. If you have time, plug these values into the original equation to check that your values are correct.
.
.
.
.
.

I'll assume you know how to differentiate a polynomial.

Gradient is zero at (2,1):
2 a x + b = 0 at x=2 --> 4 a + b = 0 ( Eq 1)

Curve passes through (1,2):
a x^2 + b x + c = y ; x=1; y=2 --> a +b +c =2 (Eq 2)

Curve passes through (2,1):
a x^2 + b x + c = y ; x=2; y=1 --> 4 a +2 b +c =1 (Eq 3)

If you need help solving three simultaneous linear equations,
ask again. I'll just mention a trick here that let me solve it in my head:
Look for common subterms in the three equations. I saw
4a + b and 4a + 2b : close! So I subtracted the Eq 1 from Eq 3 and got b+ c =1. There is a b+c in equation 2. Hurrah! Since b+c is 1 and a + b + c is 2, then a must be 1.
Then go back to eq 1 and solve for b, then go to either equation and get c.



The solution is:

a=1
b=-4
c=5

Plug those values back in to check: a x^2 + b x + c
= x^2 - 4 x + 5

At x=1, this is 2. (1,2) is on the curve - check
At x=2, this is 1. (2,1) is on the curve -check.
Differentiate. dy/dx= 2x - 4. At x=2, this is indeed zero. Good!



Hope this helps.
reilly

Answer 2

Basically you need to use the information provided to get the values of constants a, b, and c. To do this, look at the first condition given:

The curve with equation y=ax^2 + bx + c passes through the point (1,2).
Substitute x with 1 and y with 2 into the equation, you'll have:
2=a+b+c (Eq. 1)

Now, The gradient of the curve is zero (0) at the point (2,1). Differentiate the equation so it becomes dy/dx=2ax+b. dy/dx = 0 when gradient is zero, so
2a(2) + b =0 (Eq. 2)

Don't forget that (2,1) gives you an extra set of points, sub them in:
1=4a+2b+c (Eq.3)

Solve the three simultaneous equations and you'll have a,b,and c.

Answer 3

Put the point (1,2) into the equation and you have:
2=a+b+c (i)

Differentiate the equation with respect to x to find the gradient:
dy/dx=2ax+b

When that equals zero and you are at the point (2,1):
0=4a+b (ii)

At the same time the point (2,1) lies on the curve, therefore:
1=a*2^2+b*2+c=4a+2b+c (iii)

Solve the system of equations (i), (ii) & (iii):
2=a+b+c
0=4a+2b
1=4a+b+c

You can solve that.

Answer 4

uhhhhh. nup