Question

Find the parabola with equation \(y = a{x^2} + bx + c\) whose tangent line at \(\left( {1,1} \right)\) has equation \(y = 3x - 2\).

Short answer

The equation of a parabola is \(y = 2{x^2} - x\).

Step-by-step solution

Parabolas

A parabola is a2-dimensional curve that can be represented on a Cartesian plane and is mirror-symmetrical which looks very much similar to aU-shaped curve.

Evaluating tangent equations using given data.

Thegiven equation is\(y = a{x^2} + bx\).

At a point \(\left( {1,1} \right)\), we have:

\(\begin{align}y\left| {_{\left( {1,1} \right)}} \right. &= a{x^2} + bx\\a + b &= 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,.......................\left( {{\rm{eq}}.1} \right)\end{align}\)

Differentiating with respect to\(x\)as:

\(y' = 2ax + b\)

Now, the equation of a tangent line is \(y = 3x - 2\) hasslope 3, so we have:

\(\begin{align}y'\left( x \right)\left| {_{x = 1}} \right. &= 3\\2a + b &= 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..........\left( {{\rm{eq}}.2} \right)\end{align}\)

Solving for constants.

Subtract eq. (2)from (1), we have:

\(\begin{align}a + b - 2a - b &= 1 - 3\\ - a &= - 2\\a &= 2\end{align}\)

Use the value of \(a\) to find the value of \(b\).

\(\begin{align}2\left( 2 \right) + b &= 3\\4 + b &= 3\\b &= - 1\end{align}\)

The obtained constant is \(a = 2\) and \(b = - 1\).

So, the equation for parabola obtained is \(y = 2{x^2} - x\).

Hence, this is the required answer.