ASAP DERIVATIVE

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lilgnome57

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Sep 6, 2010
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find values of a and b so that:
y=ax^2 + bx
has a tangent line at (1,1) whise equation is:
y=3x-2
please explain how!
 


Substitute coordinates of tangent point into the quadratic equation.

Doing that yields an equation in a and b.

Find the derivative of ax^2 + bx at x = 1 because that's the slope of the line.

Set the derivative equal to 3.

Doing that yields a second equation in a and b.

Solve the system of two equations in a and b.

 
why do you set the derivative equal to three?
and
so the information given of the derivative (y=3x-2) is useless when solving this?
Thank You!
 
\(\displaystyle Here \ is \ the \ graph \ of \ f(x) \ = \ ax^2+bx \ and \ y \ = \ 3x-2, \ can \ you \ take \ it \ from \ here?\)

[attachment=0:22k1a11p]bbb.jpg[/attachment:22k1a11p]
 
lilgnome57 said:
why do you set the derivative equal to three?
and
so the information given of the derivative (y=3x-2) is useless when solving this?
Thank You!
Click to expand...

Can you answer these q's?
 
BigGlenntheHeavy said:
\(\displaystyle What \ is \ the \ slope \ of \ f(x) \ at \ (1,1)?\)
Click to expand...

1, when plugged into the derivative that is given to me. So why is it set equal to 3?
 
but if you plug x=1 into the derivative given to find the slope at the point:
y=3(1)-2 =1. How do you get 3?
 
unless yo take the derivative of y=3x-2 and get y=3? which makes sense. I was thinking that equation was the derivative but it is not. Thank You!
 
\(\displaystyle f(1) \ = \ 1 \ = \ a+b\)

\(\displaystyle f'(1) \ = \ 3 \ = \ 2a+b\)

\(\displaystyle Now, \ solve \ the \ system.\)
 
lilgnome57 said:
so a=2, b=-1?
Click to expand...

Check and see whether or not those parameters work.

By the way, the derivative IS the slope of a tangent line.

The slope of the tangent line is obviously 3.

That's why the derivative equals 3, in this exercise.

If the tangent line were y = 4x - 2, then the derivative would equal 4.

 
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