+118723 $$\\f(x)=2x^2-4x+9\\
f(x)=2\times x^2-4\times x+9\\\\
$for f(3) you have to put a 3 everywhere that there was an x$\\
f(3)=2\times 3^2-4\times 3+9=2\times 9-12+9=18-12+9=15\\\\
$for f(-3) you have to put a -3 everywhere that there was an x$\\
f(-3)=2\times (-3)^2-4\times (-3)+9=2\times 9+12+9=18+12+9=39\\
so\\
2f(3)+3f(-3) = 2*15+3*39 = 30+117 = 147$$
+130511 2f(3) says to put 3 into the function and multiply the result by 2
3f(-3) says to put -3 into the function and multiply the result by 3
Then just add the two above results together
Post back if you have trouble with this.....
+118723 $$\\f(x)=2x^2-4x+9\\
f(x)=2\times x^2-4\times x+9\\\\
$for f(3) you have to put a 3 everywhere that there was an x$\\
f(3)=2\times 3^2-4\times 3+9=2\times 9-12+9=18-12+9=15\\\\
$for f(-3) you have to put a -3 everywhere that there was an x$\\
f(-3)=2\times (-3)^2-4\times (-3)+9=2\times 9+12+9=18+12+9=39\\
so\\
2f(3)+3f(-3) = 2*15+3*39 = 30+117 = 147$$
