Question

For \(y=x(c-x)\) \(\frac{d y}{d x}=c-2 x\) For \(y=x^{2}+a x+b\) \(\frac{d y}{d x}=2 x+a\) At \((1,0)\) \(0=(C-1) \Rightarrow C=1\) \(0=1+a+b, 1-2=2+a\) \(\Rightarrow \quad a=-3\) \(\Rightarrow b=2\)

Short answer

Answer: a = -3, b = 2

Step-by-step solution

Find the value of 'c' at (1,0) in the first equation

Since the point \((1,0)\) lies on the first equation, we can use it to solve for \(c\). Substitute \(x = 1\) and \(y = 0\) into the first equation: \(0 = 1(c-1)\) \(0 = c - 1\) \(c = 1\)

Equate the first derivatives of both equations

At the point \((1,0)\), equate the first derivatives of the equations: \(c-2(1) = 2(1)+a\) \(1-2 = 2+a\)

Solve for 'a'

From Step 2: \(-1=2+a\) \(a=-3\)

Find the value of 'b' using the second equation at (1,0)

Since the point \((1,0)\) lies on the second equation, we can use it to solve for \(b\). Substitute \(x=1\) and \(y=0\) into the second equation and use the found value of \(a\): \(0 = (1)^2 + (-3)(1) + b\)

Solve for 'b'

From Step 4: \(0 = 1 - 3 + b\) \(b = 2\) Now we have the constants \(a\) and \(b\). The resulting equation is: \(y = x^2 - 3x + 2\)