Question look-over

[alexandra.vo]

Are you able to revise over my answer to my question, is it correct?

The curve with equation y=ax^2+bx has a gradient of -2 at the point (1,2). The values of a and b are:

Working out:
y=y1+b(x-x1)
y=2+-2(x-1)
y= -2x+4

Therefore: a= 4, b= -2

I didn't use the equation given in the question to work out the question. Is my answer wrong?

 

[HallsofIvy]

Saying that a= 4 and b= -2 means that \(\displaystyle y= ax^2+ bx= 4x^2- 2x\). When x= 1 y= 4(1)- 2(1)= 2. The derivative is \(\displaystyle y'= 8x- 2\). When x= 1 the derivative is 8(1)- 2= 6, NOT -2!

You use the equation of a straight line through (1, 2) with slope -2 so I assume you intend that as the equation of the tangent line to the parabola. But how did you decide which was "a" and which was "b"?

 

[Steven G]

Are you able to revise over my answer to my question, is it correct?

The curve with equation y=ax^2+bx has a gradient of -2 at the point (1,2). The values of a and b are:

Working out:
y=y1+b(x-x1)
y=2+-2(x-1)
y= -2x+4

Therefore: a= 4, b= -2

I didn't use the equation given in the question to work out the question. Is my answer wrong?

The equation of a line can be written in the form y = ax + b (or y = bx + a)/
The equation of a quadratic can be written in the y = ax^2 + bx +c.

Why would you assume that the two a's are the same and that the two b's are the same??

 

[alexandra.vo]

I simply thought that it was correct using the form y=ax+b and then take the values for a and b.

I calculated the derivative as y= 2ax+b, then subbed the points in.

Am I correct as my answer being a= -4 and b= 6?