Fun Calculus Problem

[Panphobia]

So on my test there were a couple pretty fun questions

code:

There is a function y = ax^2 + bx + c , if the x intercepts are 0 and 8, and the slope of the tangent at x =2 is 16, what are a, b, c

I got a = -4, b = 32, c =0. The second one

code:

The function y=x^3+3 has a tangent at (1,4) this tangent also intersects at another point P, find point P

that one was pretty easy but fun, my friend was convinced that I got it wrong, but I am pretty sure I am right saying that, P = (-2,-5) because this is what I did

code:

The derivative function d/dx = 3x^2 so the slope of the line is 3(1)^2, then we know that y = mx + b so you sub (1,4) into that, and also you sub the slope, so 4 = 3 + b, and b = 1, so y = 3x + 1, now to get the intersection point you make y = y, 3x + 1 = x^3 + 3 0 = x^3 - 3x + 2 0 = (x-1)(x-1)(x+2) since we already know x=1 is an intersection point we find that -2 is the only different one so we sub -2 into y = x^3 +3 and get y = -5 so P = (-2,-5)

am I right?

[Raknarg]

Yeah I got the same answers for both

[Panphobia]

Ok so I wrote another calculus test recently, and the thinking question was similar to that of the previous here it is

code:

The function f(x) = ax^3 + bx^2 + c goes through point (2,-3) and has an inflection point at (1,1),with this information what are the values of the constants a,b,c

I got answers that

code:

a = 2, b =  -6, and c = 5

I was just wondering if I got it right, thank you and that is all.

[d310]

I agree with the answer.

[Panphobia]

How the hell would I answer a question like this

code:

Consider the function f(x) = ax^3 + bx^2 + cx + d what is the equation to the tangent of the only inflection point?

How would I get one single answer for this? Without it being crazy with variables all over?

[Dreadnought]

Well first you solve for the inflection point (you should know how to do this).

Then you find the slope of the tangent at that point. You probably also know that the slope is the first derivative evaluated at the inflection point.
So substitute in the value of x at the inflection point and simplify it into something relatively nice (2 terms).

Then finding the equation of the tangent line should be easy (you know the slope and the y-intercept is pretty obvious).

[Panphobia]

I mean, I know how to solve it easily with all those variables in the equation, but I mean to solve for a definite line, is that possible?

[Dreadnought]

Well the line does depend on the values of a,b,c, and d so you can't get a single line for all values (that would imply all tangents to a cubic are a single line). But you can give the tangent line in a form that depends on a,b,c and d. Then if you knew a,b,c and d you would just substitute in the values and have a nice equation.

[Panphobia]

yea thats what I did, because it said find an equation of the tangent to that function. The question on the test was worded weird but yea, I did it that exact way. and the equation of the tangent to the inflection point does not depend on d, only a, b and c . This is because when you take the derivative(the tangent to any point on the graph), the d value disappears because it is like f(x) = dx^0 f'(x) = 0dx^-1

[Dreadnought]

The slope of the tangent line does not depend on d but the equation does. Remember that to define a line in the Cartesian plane, you need both a slope and a point on the line (and this point will depend on d).

EDIT:
Maybe it's this point that's bugging you.
You might be used to giving the equation of the tangent line in the form y = mx + b (m is the slope, b is the y intercept), however b isn't necessarily nice in this case. So I give you the linear approximation to f at the point a. L(x) = f'(a)*(x-a) + f(a) where a is point around which you want to approximate (in this case the inflection point).

L(x) is the tangent line to f(x) at the point x = a.
Does this make it easier?

EDIT to the edit: You'll notice that this is actually what you would get if you solved for b and substituted into y = mx + b.
EDIT 3: Ok I'm feel I need to show you where L(x) comes from.

code:

You want a line with a slope of f'(a) going through f(a) where a is the inflection point. We find the y-intercept of this line, b = f(a) - f'(a)*a   <--- comes from f(a) = f'(a)*a + b since f(a) is on the line y = f'(a)*x + b Substitute into L(x) = f'(a)*x + b L(x) = f'(a)*x + f(a) - f'(a)*a L(x) = f(a) + f'(a)*(x-a)

Edit 4: Does this help?

[Panphobia]

Yea I understand it, but I got the question right on the test anyway

[Dreadnought]

Oh, ok well good then! (I needed something to distract me from boring homework anyway)

[Panphobia]

this unit was probably the unit where I had to do some work, next unit is optimization, I think it is probably going to be the easiest unit of the semester have you taken it already?

[Dreadnought]

Oh yes I've taken it, a couple years ago... I recently finished second year math at U of Waterloo.

Optimization is a great application of the calculus you learn in high school. It's much easier than anything you've learned prior (like completing the square to find the vertex of a parabola).

Considering you were asking about a tricky differential equation the other day I'd say you should be fine for optimization. (if you teacher gave you that question on a test I find that kinda evil...)

[Panphobia]

my teacher is evil, he puts like university integration questions on the tests, when in reality integration isn't a part of the course. Anything that is not in the course syllabus you are not allowed to assess on. Sooooooo,....