$f(x)=2x^2-3x+4$

$f^{\text{'}}(x)=4x-3$

$f^{\text{'}\text{'}}(x)=4$

USE TRE SECOND DERIN.

to DETERMIAE WHETHR

$,$ POSITIVE, NEGATIVE, OR ZORO

YOINT IN THE TABLE.

$\backslash$table$[[x,f(x),f^{\text{'}\text{'}}(x),],[-2,18,,],[-1,9,,],[0,4,,],[1,3,,],[0,2,6,],[7,3,13,]]$

THEN DETERMINE THE INTERVAL$($S$)$ OVER WHICH $f(x)$ is CONCAVE UP AND OVER WHICH $f(x)$ is CONCAVE DOWN.

Solve $29$ and $33$ please