# Math Help - Graphing the curve and both tangents...

1. ## Graphing the curve and both tangents...

QUESTION:

(a) Find the slope of tangent to the curve y=3+4x^2-2x^3 at the point where x=a.
(b) Find equations of the tangent lines at the points (1,5) and (2,3).
(c) Graph the curve and both tangents on a common screen.

(a) m(TL) = dy/dx = 8x-6x^2

(i) at P(1,5); m(TL) = 2
(ii) at P(2,3); m(TL) = -8

(b)

(i) y = 2x +3
(ii) y = -8x+19

(c) now in this part of graphing i don't know how to graph the curve since it is raised to cube..i don't know either what kind of curve is this..

^_^

2. A method that always works is to put in points for the x to solve for the y to plot on your graph.
Eg.
If x = -0.5, then y = 4.25
If x = 0, then y = 3
If x = 0.5, theny = 3.75
And then join up the dots

Here's a common image of a cubic graph generated by Wolfram|Alpha:

Your graph should look something like this one, maybe moved up a bit.
Then just add the tangent lines to your graph for the 2 tangent lines.

3. oh yeah!.. like test points.. thanks a lot ^_^

4. Originally Posted by cutiemike1
QUESTION:

(a) Find the slope of tangent to the curve y=3+4x^2-2x^3 at the point where x=a.
(b) Find equations of the tangent lines at the points (1,5) and (2,3).
(c) Graph the curve and both tangents on a common screen.

(a) m(TL) = dy/dx = 8x-6x^2
Strictly speaking, this is wrong. The problem asked for the derivative " at the point where x=a." The correct answer would be $dy/dx= 8a- 6a^2$.

(i) at P(1,5); m(TL) = 2
(ii) at P(2,3); m(TL) = -8
Except to solve part (b) there were no parts "i" and "ii". You were not asked for the slope of the tangent at x= 1 and x= 2 in part (a),

(b)

(i) y = 2x +3
(ii) y = -8x+19

(c) now in this part of graphing i don't know how to graph the curve since it is raised to cube..i don't know either what kind of curve is this..

^_^

5. 1. So does that mean, i cannot get the slopes given two points, it's because of "x=a"?
2. It is just a guide for me to know where the slope is leaning.