Question

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Answer 1

$\huge{\underline{\underline{\mathcal\color{blue}{\bigstar{Answer}}}}}$

$\red{y=2x^2-3x+1}$

Step-by-step explanation:

$\rightsquigarrow$ So we want to find the equation of a parabola of the form:

$\bold{y=ax^2+bx+c}$

$\rightsquigarrow$ This passes through (0,1) and is tangent to the line y=x-1 at (1,0).

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

$\bold{y=ax^2+bx+1}$

$\rightsquigarrow$ Now, look at the equation of the tangent line. The slope of the tangent line is 1. In other words, the derivative of our parabola at x=1 must be 1.

So, differentiate our equation:

$\bold{y=ax^2+bx+1}$

$\rightsquigarrow$ Differentiate:

$\bold{y'=\dfrac{d}{dx}[ax^2+bx+1]}$

$\rightsquigarrow$ Expand:

$\bold{y'=\dfrac{d}{dx}[ax^2]+\dfrac{d}{dx}[bx]+\dfrac{d}{dx}[1]}$

$\rightsquigarrow$ Power Rule. Since we're differentiating with respect to x, treat a and b as constants. So:

$\bold{y'=2ax+b}$

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

$\bold{1=2a(1)+b}$

$\rightsquigarrow$ Remember that the derivative gives you the slope of the line tangent at a certain point. Since the slope of the tangent line at x=1 is 1, this means that our derivative when x=1 must be 1.

$\rightsquigarrow$ Simplify:

1=2a+b

$\rightsquigarrow$ 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 1 if we substitute in zero. So:

$\tt{y=ax^2+bx+1}$

$\rightsquigarrow$ Substitute 0 for y and 1 for x. Therefore:

$\tt{0=a(1)^2+b(1)+1}$

$\rightsquigarrow$ Simplify:

$\tt{0=a+b+1}$

$\rightsquigarrow$ Now, we have a system of equations. Let's solve for both a and b.

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

$\begin{gathered}1=2a+b\\1-2a=b\end{gathered}$

$\rightsquigarrow$ Substitute this into the equation we just acquired:

$\tt{0=a+(1-2a)+1}$

$\rightsquigarrow$ Solve for a. Combine like terms:

$\bold{0=(a-2a)+(1+1)}$

$\rightsquigarrow$ Add or subtract:

0=-a+2

$\rightsquigarrow$ Add a to both sides:

$\therefore{\boxed{\purple{a=2 }}}$ 

$\rightsquigarrow$ Therefore, the value of a is 2.

To find b, substitute it into 1-2a=b:

1-2(2)=b

Multiply:

1-4=b

Subtract:

$\therefore{\boxed{\purple{ b=-3  }}}$ 

$\rightsquigarrow$ Therefore, let's substitute them into our very first equation to find our final answer:

$\tt{y=ax^2+bx+1}$

$\hookrightarrow$ Substitute a for 2 and b for -3. Therefore, our solution is:

$\bold{y=2x^2-3x+1}$