Curve y=ax^2+bx+c passes through (1,2) and is tangent to y=x at the origin find (a,b,c)?? i have the answer i have forgotten how did i solve it, Someone please explain in a little detail. Thanks

Question

sunnnystrong · Answer

Since the curve passes through the point (1,2), we can start by substituting that into the equation

$$2=a(1)^2+b(1)+c$$

$$2=a+b+c$$
Next, we know that y=x is tangent to the curve at the origin
Slope of the line is 1-->

$$\frac{ dy }{ dx }=2ax+b$$
@ the origin -->

1=2(a)(0)+b
1=b.

Evaluate y @ (0,0)

$$0=a(0)^2+b(0)+c$$
0=c

Find a-->
$$2=a(1)^2+1(1)$$
a=1


.... @hafeda so what is the equation of the curve?

hafeda · Answer

Thank u so much, what equation are u asking for ???

sunnnystrong · Answer

NP Always happy to help :D
$$ y=ax^2+bx+c $$

.. if you know a,b,c whats the curve?

hafeda · Answer

ohh y=x^2+x is the curve