Tanget lines and points and of graphs

[intervade]

Ok, I have a couple of questions that I'm completely lost on.

1) Find values of a and b so that y = ax^2 + bx has a tanget line at (1,1) whose equation is y=3x-2

I basically have my derivative of 3(I think) so where do I go from there?

2) Find the equation of the line(s) that are tanget to the graphs of y = x^2 and y = -x^2 + 6x - 5.
This one leaves me clueless

 

[galactus]

You must find the derivative of the quadratic, not the line. The line equation does tell us that the slope at that point is 3.

So, by \(\displaystyle y'=2ax+b\Rightarrow 3=2a+b\)

Also, \(\displaystyle 1=a+b\), since it is tangent at (1,1).

So, there are two equations to solve for a and b and we have it.

\(\displaystyle 2a+b=3\)

\(\displaystyle a+b=1\)

Solve away.


For the second problem, see here:

viewtopic.php?f=3&t=17238&p=68276

 

[intervade]

Hmmm I understand the derivative of the quadratic but where did you get the 3 = 2a + b and the other equation , 1 = a + b?

 

[mmm4444bot]

… I understand the derivative of the quadratic but where did you get the 3 = 2a + b …

Galactus clearly showed you.

This is given: y ' = 3 at x = 1

You state that you understand: y ' = 2ax + b

Substitute the given 3 for y ' and substitute the given 1 for x into this equation that you understand.