#### macewan3135

I am reviewing for a midterm and i am stuck on the problem that follows.

Evaluate f(-3), f(0), f(2), for the piecewise defined function. Then sketch the graph of the function.

f(x)= -1 if x</= 1
f(x)= 7-2(x) if x > 1

I plugged it all in and came up with up
f(-3)=-1
f(0)= -1
f(2) = 7-2(2) = 3

I am stuck on how to graph it, I am looking at a different example and I am not sure where they get the points or shape from. Any help would be greatly appreciated.

#### skeeter

MHF Helper
graph it in pieces since it's a piece-wise defined function ...

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#### macewan3135

Where do you get the points from? And what does the empty dot compared to solid dot represent? I have forgot everything I learned last year.

#### skeeter

MHF Helper
I didn't use points ... I graphed each piece of the function separately

$f(x) = -1$ ... a horizontal line at $y = -1$ that starts at $x = 1$, includes $1$, and goes to the left since the function is valid for values $x \le 1$

f(x) = 7-2x ... a line with slope, $m=-2$, y-intercept $y=7$, that starts at $x = 1$, but does not include $x = 1$ since the defined interval is $x > 1$

This was an extremely easy one, they will get more difficult. My advice is to review, or learn it if you did not in precalculus ...

[video]https://youtu.be/-gwffMEr8i8[/video]

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#### macewan3135

Just last question I am confused because you say Y intercept = 7 but you don't cross it on your graph you attached?

#### skeeter

MHF Helper
That's right ... the y-intercept is located where x = 0, but the function is only defined for values of x > 1.

If you extend the line to the left (in your head), it will pass thru (0,7) ... that's why I included the grid.

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#### macewan3135

Thank you so much. I was over thinking the (Swear) out of it. Life Saver Last edited by a moderator: