How to solve the unknown variable using calculus?

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Can someone please explain to me how to do question 14? Thanks!

2 Answers
Sep 23, 2017

See below.

Explanation:

The tangent line is given by

L->p = Q+lambda(U-V) with p = (x,y) and given

f(p)=y-(ax^2+bx) we have also

f(p)=f(Q+lambda(U-V))=0 or solving for lambda

lambda = (4 a u + b u + v pm sqrt[8 a u (2 u + v) + (b u + v)^2])/(2 a u^2)

but if L is tangent to f then

8 a u (2 u + v) + (b u + v)^2=0

then solving the system

{(8 a u (2 u + v) + (b u + v)^2=0),(f(Q)=0):}

at u=6 we obtain

a = -1/12 (12 +v), b = (24 + v)/6

but

-u/v = -6/v = dy/dx at Q then

-6/v = 2a x+b=4a+b and then solving

{(a = -1/12 (12 +v)),(b = (24 + v)/6),(-6/v=4a+b):}

we get the solution.

Sep 23, 2017

Please see below.

Explanation:

y = ax^2+bx

From the fact that #(2,4) lies on the parabola, we get

4 = a(2)^2+b(2), so

color(blue)(2a+b=2) " " eq (1)

The slope of the tangent line can be found in two ways

Using the definiton of the slope of the tangent line (the derivative) we get:

dy/dx = 2ax+b

So at x=2, we have m = 4a+b

AND

we can find the slope using the two points (2,4) and (6,0)

m = (0-4)/(6-2) = -1

This gives us

color(blue)(4a+b=-1) " " eq (2)

Now solve the system formed be eq (1) and eq (2).

{(2a+b=2),(4a+b=-1) :}

Subtracting the first from the second gets us:

2a = -3, so a = (-3)/2

And the first equation now gets us 2((-3)/2)+b = 2 so

-3+b = 2 and b = 5.