Question

Hello! Please explain how to solve this step-by-step. (Question 111.) I tried solving but not sure if I did it correctly. Thanks!

Answer 1

`y=2x^2-3x+1`

Explanation:

We want to find the equation of a parabola of the form:

`y=ax^2+bx+c`

This passes through (0,1) and is tangent to the line:

`y=x-1`

At the point (1,0).

First, note that (0,1) is the y-intercept. In other words, our constant c is 1. Thus:

`y=ax^2+bx+1`

From the equation of the tangent line, we can see that it has a slope of one.

Recall that the slope of the tangent line at a point is equivalent to the value of the derivative at the same point.

In other words, the value of the derivative of our parabola at x = 1 must be one.

Find the derivative. Take the derivative of both sides with respect to x:

`\displaystyle (dy)/(dx)=(d)/(dx)\left[ax^2+bx+1\right]`

Expand:

`\displaystyle (dy)/(dx)=(d)/(dx)\left[ax^2\right]+(d)/(dx)\left[bx\right]+(d)/(dx)\left[1\right]`

Use the Power Rule. Since we're differentiating with respect to x, we can treat a and b as constants. Thus:

`\displaystyle (dy)/(dx)=2ax+b`

Now, since the slope of the tangent line at x = 1 is 1, this means that:

`(1)=2a(1)+b`

Simplify:

`1=2a+b`

Let's hold on to this equation for now.

Since the line is tangent at the point (1,0), this means that our original function equals zero when x = 1. In other words:

`0=a(1)^2+b(1)+1`

Simplify:

`0=a+b+1`

This yields the following system of equations:

`\displaystyle \begin{cases} 2a+b=1\\ a+b+1=0\end{cases}`

Solve for a and b.

From our previous equation, let's subtract 2a from both sides:

`b=1-2a`

Substitute this into the newly acquired equation:

`0=a+(1-2a)+1`

Solve for a. Rewrite:

`0=(a-2a)+(1+1)`

Combine like terms:

`0=-a+2`

Hence:

`a=2`

Find b:

`b=1-2(2)=-3`

Therefore, a = 2 and b = -3.

Then by substitution, we can see that our final equation is:

`y=2x^2-3x+1`