Use both coordinates and gradients at stationary points

Let $f(x)=Ax^3+Bx^2+Cx+D$, so $f\prime(x)=3Ax^2+2Bx+C$. The source callout says stationary points imply zero derivative. Therefore $12A-4B+C=0$ and $\tfrac34A+B+C=0$.

Stationary points also lie on the curve, as the second source callout stresses. Their ordinates give $-8A+4B-2C+D=8$ and $\tfrac18A+\tfrac14B+\tfrac12C+D=\tfrac{189}{8}$. Together these four linear equations determine the four coefficients.

The printed calculator screen gives $x_1=-2$, $x_2=-9/2$, $x_3=6$, $x_4=22$. Its callout maps these to $A,B,C,D$ respectively and warns against copying the screen variable names into the final solution. Thus $f(x)=-2x^3-\tfrac92x^2+6x+22$. Verify $f(-2)=8$, $f(1/2)=189/8$ and $f\prime(-2)=f\prime(1/2)=0$.