Question

g(x) = 4x - x^2
Find the equations of the two tangent lines to the graph of g that pass through P(2,5), which is not on the graph g(x)

Answer 1

draw a graph of g(x)

should g not be
g(x) = 4x² − x
or something, because otherwise the question makes no sence (g would be a straight line)

Answer 2

Nice catch. Edited.

Answer 3

Hang on, I'm in the middle of answering another question

Answer 4

Ok, we have
g(x) = 4x - x^2
Find the equations of the two tangent lines to the graph of g that pass through P(2,5), which is not on the graph g(x)

Just a sec, need a spreadsheet to create a nice graph
hang on

Answer 5

Can I do this without a graph?

Answer 6

Preferably with the first derivative.

Answer 7

Yest that's right

Answer 8

The tangent lines have the direction of the derivative of g at some point on g

Answer 9

Here is what I got

the tangents have a direction of dg/dx and are straight lines

y = ax + b
where a is dg(x)/dx | x on g and x on the tangent line as well


g'(x) = 4 - 2x_g, x_g is an x on the parabola x

so
y_t = (4 - 2x_g)x_t + b

the tangent line passes through (2, 5)
5 = (4 - 2x_g)2 + b
5 = 8 -4x_g + b
b = 4x_g - 3

There is also a point on the tangent line and the parabola:
(x_t, y_t) = (x_g , y_g),

y_t = (4 - 2x_g)x_t + b
y_g = 4x_g - x²_g

x_g = x_t and y_t = y_g, so
(4 - 2x_g)x_g + b = 4x_g - x²_g
4x_g -2x²_g +b = 4x_g - x²_g
b = x²_g

We already had an expression for b so
4x_g - 3 = x²_g
x²_g - 4x_g +3 = 0

so x is
(4 + sqrt(16 - 12))/2 = 3
and
(4 - sqrt(16 - 12))/2 = 1

so the first tangent line is
y = (4 - 2x_g)x + x²_g , x_g = 3
y = (4 - 2*3)x + 9
y = -2x + 9

and the second tangent line is
y = (4 - 2x_g)x + x²_g , x_g = 1
y = 2x + 1

Answer 10

Yes you can do it without a graph, but a graph shows what is going on
and shows the anwers make sense

A sketch in advance is also handy



I usesed this site to make the graphs
http://rechneronline.de/function-graphs/