Problem source

Find the coordinates of the turning point on the curve $y = x^2 - 2bx + c\,$. Sketch the curve in the case that the equation $x^2 - 2bx + c=0$ has two distinct real roots. Use your sketch to determine necessary and sufficient conditions on $b$ and $c$ for the equation $x^2 - 2bx + c = 0$ to have two distinct real roots. Determine necessary and sufficient conditions on $b$ and $c$ for this equation to have two distinct positive roots.   
   
Find the coordinates of the turning points on the curve $y = x^3 - 3b^2x + c$ (with $b>0$) and hence determine necessary and sufficient conditions on $b$ and $c$ for the equation $x^3 - 3b^2x + c = 0$    
to have three distinct real roots. Determine necessary and sufficient conditions on $a\,$, $b$ and $c$ for the equation $\l x - a \r^3 - 3b^2 \l x - a \r + c = 0$ to have three distinct positive roots.   
   
Show that the equation $2x^3 - 9x^2 + 7x - 1 = 0$ has three distinct positive roots.

Solution source

\begin{align*}
y &= x^2-2bx+c \\
&= (x-b)^2+c-b^2
\end{align*}

Therefore the turning point is at $(b,c-b^2)$

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)^2-2*3*(#1)+1};
    \def\xl{-2};
    \def\xu{8};
    \def\yl{-10};
    \def\yu{10};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
    \end{scope}

    \filldraw (3,-8) circle (1pt) node[below] {$(b,c-b^2)$};
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Therefore there will be two distinct roots if $c -b^2 < 0 \Rightarrow c < b^2$.

For the equation to have two distinct positive roots it needs to have two distinct roots (ie the condition above) and $y(0) = c$ needs to be positive, ie $c > 0$. Therefore the curve has two distinct positive roots if $0 < c < b^2$.

The turning points on $y = x^3-3b^2x+c$ will have $0 = y' = 3x^2-3b^2 \Rightarrow x = \pm b$

Therefore for the cubic to have three distinct real root we must have a root between the turning points, $y(-b) > 0 > y(b)$

$b^3-3b^3+c = c-2b^3 < 0$
$(-b)^3+3b^3+c = c+2b^3 > 0$
ie $-2b^3 < c < 2b^3$.

The equation  $\l x - a \r^3 - 3b^2 \l x - a \r + c = 0$  will have 3 distinct roots if

$-2b^3 < c < 2b^3$, they will all be positive if the $y(0) < 0$ and $a+b > 0$ (ie the first turning point is in the first quadrant, ie $-a^3+3b^2a+c < 0, a+b>0$.

\begin{align*}
2x^3 - 9x^2 + 7x - 1 &= 2(x^3-\frac92x^2+\frac72 x-\frac12) \\
&= 2((x-\frac{3}{2})^3-\frac{27}{4}x+\frac{27}{8}+\frac72x-\frac12) \\
&=  2((x-\frac{3}{2})^3-\frac{13}{4}x+\frac{23}{8})  \\
&=  2((x-\frac{3}{2})^3-\frac{13}{4}(x-\frac{3}{2})-\frac{39}{8}+\frac{23}{8})  \\
&=  2((x-\frac{3}{2})^3-\frac{13}{4}(x-\frac{3}{2})-2)  \\
\end{align*}

Therefore in our notation $a = \frac32, b = \sqrt{13/12}, c = 2$.

Clearly $a+b > 0$, so we need to check $|c| < 2b^3$ which is clearly true as $b > 1$.

Finally we need to check $y(0) = -1$, so all conditions are satisfied and there are 3 distinct roots. 

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)^3-9*(#1)^2+7*(#1)-1};
    \def\xl{-2};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
    \end{scope}

    % \filldraw (3,-8) circle (1pt) node[below] {$(b,c-b^2)$};
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}