Here is another problem from me.
I tried solving this, and I am thinking that I somehow need to turn this into an equation with some known points and than solve for a,b and c as a simultaneous equation. However, whenever I try to do this I get lost.
The curve y = ax2 + bx + c passes through the point (1; 2) and is tangent to the line
y = x at the origin. Find a, b and c
Thanks to everyone who reads this
"The curve y = ax² + bx + c is tangent to the line y = x at the origin"
To be tangent at the origin, at least the curve must pass through the point (0; 0). So 0=a^0+b*0+c. Hence c=0
The slope is y'(x)=2ax+b. The slope of the line y=x is 1. So, at the origine, 1=2*0+b. Hence b=1
y = ax²+1
The curve passes through the point (1; 2). So, 2=a*1²+1. Hence a=2-1=1
y = x²+x