More help for Differentiation , questions with diagrams:> (1 Viewer)

[HKHSCstudent]

:wave: :rofl:
Please help me, sorry the diagrams are a bit out of shape, hard to draw in the computer

1.The diagram shows the graph of a ctertain function y =f(x). The graph has a maximum turning points and a point of horizontal inflexion in the domain shown.

On the same set of axes, draw a sketch of the derivative f''(x) of the function.


2. The diagram shows a graph of the parabola y+x^2 and the tangent to the parabola at x=a
1) Find the gradient of the parabola at x=a.
2) Find the equation of the tangent at x+a.
3) Find the value of a if the tangent intersects the x-axis at x=2

Thanks a lot for those who help me !

 

[flier higher]

2. The diagram shows a graph of the parabola y+x^2 and the tangent to the parabola at x=a
1) Find the gradient of the parabola at x=a.
2) Find the equation of the tangent at x+a.
3) Find the value of a if the tangent intersects the x-axis at x=2

sorry 'bout rmv the smilies and not answering q1; i don't know how i'm gonna draw it and have it appear here

2. info:
parabola: y=x^2 (i'm assuming that it is. it's concave up in the diagram. is it y=x^2?)
tangent at x=a

pt.1: to find m at x=a, differentiate eqn y=x^2
dy/dx = 2x

then sub value (x=a) in

.: dy/dx = 2a

pt 2. i'm guessing that's x=a, not plus.

using the point-gradient form, y-y₁=m(x-x₁), into which you sub your m and y1 and x1,
y-a^2=2a(x-a) (a^2 is your y-value at x=a, using your eqn for the curve.)
y-a²=2ax-2a²
.: eqn of tangent is y+a^2-2ax=0

part. 3
since pt (2,0) lies on the line y+a^2-2ax=0, sub in ur values and solve for a:
(0)+a^2-2a(2)=0, which simplifies down to
a^2-4a=0
a(a-4)=0
a=0, 4

sub back in and check if you want.

please correct me on my working out, setting out, or whatever. cheers.
more queries?

toodles, then. =)