Question

The curve ax^2+bx+c passes through the point (1,2) and is tangent to the line y=x at the origin. Find a, b, and c.

Answer 1

Is it y=ax^2+bx+c ?

Answer 2

Correct.

Answer 3

Thoughts?

Answer 4

Substitute the point (1,2) in y=ax^2+bx+c
you will get your 1st equation

Now the curve is tangent to y=x at (0,0) So, substitute (0,0) in equation y=ax^2+bx+c
you will get a second equation

the 1st derivative of y=ax^2+bx+c i.e. y'=2ax+b at (0,0) will give you slope of tangent of the curve at (0,0).. Substituting the point (0,0) in y'=2ax+b you get slope(m)=b
but the slope of y=x is 1
so, you have b=1

So, you have b=1 and two other equations.. Solve them to get a and c..

Answer 5

Okay, between this and my friend, I've figured it out! Thank you, sir!