Answer:
[tex]y=2x^2-3x+1[/tex]
Step-by-step explanation:
We want to find the equation of a parabola of the form:
[tex]y=ax^2+bx+c[/tex]
This passes through (0,1) and is tangent to the line:
[tex]y=x-1[/tex]
At the point (1,0).
First, note that (0,1) is the y-intercept. In other words, our constant c is 1. Thus:
[tex]y=ax^2+bx+1[/tex]
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:
[tex]\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[ax^2+bx+1\right][/tex]
Expand:
[tex]\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[ax^2\right]+\frac{d}{dx}\left[bx\right]+\frac{d}{dx}\left[1\right][/tex]
Use the Power Rule. Since we're differentiating with respect to x, we can treat a and b as constants. Thus:
[tex]\displaystyle \frac{dy}{dx}=2ax+b[/tex]
Now, since the slope of the tangent line at x = 1 is 1, this means that:
[tex](1)=2a(1)+b[/tex]
Simplify:
[tex]1=2a+b[/tex]
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:
[tex]0=a(1)^2+b(1)+1[/tex]
Simplify:
[tex]0=a+b+1[/tex]
This yields the following system of equations:
[tex]\displaystyle \begin{cases} 2a+b=1\\ a+b+1=0\end{cases}[/tex]
Solve for a and b.
From our previous equation, let's subtract 2a from both sides:
[tex]b=1-2a[/tex]
Substitute this into the newly acquired equation:
[tex]0=a+(1-2a)+1[/tex]
Solve for a. Rewrite:
[tex]0=(a-2a)+(1+1)[/tex]
Combine like terms:
[tex]0=-a+2[/tex]
Hence:
[tex]a=2[/tex]
Find b:
[tex]b=1-2(2)=-3[/tex]
Therefore, a = 2 and b = -3.
Then by substitution, we can see that our final equation is:
[tex]y=2x^2-3x+1[/tex]
