Question

find the parabola with equation $y = ax^2 + bx$ whose tangent line at $(2, 8)$ has equation $y = 10x - 12$. $y = 2x^2 - 4x$

Explanation:

Step1: Use the point on the parabola

The point \((2, 8)\) lies on the parabola \(y = ax^{2}+bx\), so substitute \(x = 2\) and \(y=8\) into the equation:
\(8=a(2)^{2}+b(2)\), which simplifies to \(8 = 4a + 2b\). We can divide this equation by 2 to get \(4=2a + b\) (Equation 1).

Step2: Find the derivative for the slope

The derivative of \(y=ax^{2}+bx\) with respect to \(x\) is \(y'=2ax + b\). The slope of the tangent line at \(x = 2\) is equal to the derivative at \(x = 2\). The tangent line is \(y = 10x-12\), so its slope is 10.
So, substitute \(x = 2\) into the derivative: \(10=2a(2)+b\), which simplifies to \(10 = 4a + b\) (Equation 2).

Step3: Solve the system of equations

We have the system of equations:
Equation 1: \(2a + b=4\)
Equation 2: \(4a + b=10\)

Subtract Equation 1 from Equation 2:
\((4a + b)-(2a + b)=10 - 4\)
\(4a + b-2a - b = 6\)
\(2a=6\), so \(a = 3\).

Substitute \(a = 3\) into Equation 1: \(2(3)+b=4\)
\(6 + b=4\), so \(b=4 - 6=- 2\).

Step4: Write the equation of the parabola

Substitute \(a = 3\) and \(b=-2\) into \(y=ax^{2}+bx\):
\(y = 3x^{2}-2x\).

Answer:

\(y = 3x^{2}-2x\)