1. C3 question

f(x) = |2x + 5| for all real values of x

a) Sketch the graph y = f(x), showing the coordinates of any points where the graph meets the coordinate axes.

b) evaluate ff(-4)

g(x) = f(x + k) for all real values of x

c) state the value of the constant k for which g(x) is symmetrical about the y axis.

OK for part a i have sketched a line crossing the y axis at +5 and bouncing off the x axis at x=-2.5

Can anybody confirm this is right?

For part b i'm not sure what to do. I guess u work out f(-4)= |-8 + 5|

=|-3| = 3 ....???? then work out f(3)....????

part c im unsure of...

2. Originally Posted by djmccabie
f(x) = |2x + 5| for all real values of x

a) Sketch the graph y = f(x), showing the coordinates of any points where the graph meets the coordinate axes.

b) evaluate $\displaystyle ff(-4)$

g(x) = f(x + k) for all real values of x

c) state the value of the constant k for which g(x) is symmetrical about the y axis.

OK for part a i have sketched a line crossing the y axis at +5 and bouncing off the x axis at x=-2.5

Can anybody confirm this is right?

For part b i'm not sure what to do. I guess u work out f(-4)= |-8 + 5|

=|-3| = 3 ....???? then work out f(3)....????

part c im unsure of...

The points on the graph are correct. It would never cross the x-axis.

$\displaystyle f(x) = |2x + 5|$

$\displaystyle ff(-4):$

$\displaystyle f(f(-4)) = f( |2(-4) + 5| ) = f( |-8 + 5|) = f(|-3|) = f(3) = |2(3) + 5| = |6 + 5| = 11.$

So yes you are correct here as well.

$\displaystyle g(x) = f(x + k)$ for all real values of x

c) state the value of the constant k for which g(x) is symmetrical about the y axis.

$\displaystyle f(x) = |2x + 5|$
$\displaystyle g(x) = f(x+k) = |2(x+k) + 5| = |2x + 2k + 5|$
The answer would be $\displaystyle k = -2.5$. Try drawing that graph and you'll see it is symmetrical about the $\displaystyle y$-axis.

3. fixed it

4. Thanks a lot for your help (quite a lot today haha) just a bit unser about part C

I understand how you get |2x + 2k + 5|

But i'm not sure how you derive k=-x

5. Sorry i got that bit wrong, but i've fixed it now.

6. hi i still dont know how you get this :/ there are 2 unknows in the eaution and it is not equal to anything...

7. The only way you will have a symmetric graph of the form $\displaystyle y = |ax + c|$ is if c = 0. If $\displaystyle c \neq 0$ then it won't be symmetrical about the y-axis. So you want to work out what k to put in that will cancel out the +5.

8. THANKS! i understand it now