Question
Given f(x) as shown, graph the following: [the line is going through (-2,-3) and (2,0)]
1. f(x) + 2
==> My graph I have is a line going through the points (-2,-1) and (2,2). Is this correct?
2. 2f(x)
==> Can someone help me with this one please? I thought the 2 in front of the f(x) would affect the amplitude of the graph, but I don't see how this graph would have an amplitude.
Any help is GREATLY appreciated!!!!!
1. f(x) + 2
==> My graph I have is a line going through the points (-2,-1) and (2,2). Is this correct?
2. 2f(x)
==> Can someone help me with this one please? I thought the 2 in front of the f(x) would affect the amplitude of the graph, but I don't see how this graph would have an amplitude.
Any help is GREATLY appreciated!!!!!
Answers
jim
OK. The original function is
f(x)=(3/4)x-3/2
1. Yes, the function (3/4)x + 1/2 does pass through those points (-2,-1) and (2,2), as you can verify by plugging them in
2. You're right; there is no amplitude - but there is a slope! It doubles the slope, and the y-intercept.
2f(x) = 2(3/4)x-2(3/2)
=(3/2)x -3
That passes through (0,-3) and (-2,0)
f(x)=(3/4)x-3/2
1. Yes, the function (3/4)x + 1/2 does pass through those points (-2,-1) and (2,2), as you can verify by plugging them in
2. You're right; there is no amplitude - but there is a slope! It doubles the slope, and the y-intercept.
2f(x) = 2(3/4)x-2(3/2)
=(3/2)x -3
That passes through (0,-3) and (-2,0)
Emily
Ohhhh I get it :) I have like two more questions though --
3. f(-2x)
==> Would you multiply the x- and y- coordinates by -2? I'm soooo confused :-/
And for number 8, the graph is a parabola instead, with the vertex at (0,0), and the "arms" of the parabola going through the points (-1,1), (1,1), and (2,4). A parabola doesn't have an amplitude either, so what exactly would you do here? (don't worry -- this isn't all my homework, these are just a few examples so I know how to do the others)
3. f(-2x)
==> Would you multiply the x- and y- coordinates by -2? I'm soooo confused :-/
And for number 8, the graph is a parabola instead, with the vertex at (0,0), and the "arms" of the parabola going through the points (-1,1), (1,1), and (2,4). A parabola doesn't have an amplitude either, so what exactly would you do here? (don't worry -- this isn't all my homework, these are just a few examples so I know how to do the others)
jim
You don't multiply the co-ords, exactly. What you're scaling is the coefficients.
But careful: f(-2x) is not -2f(x)!
f(-2x) means might be a shorthand for f(g(x)), where g(x) = -2x.
-2f(x) would be
-2(3/4)x+2(3/2)
=3-(3/2)x
Your parabola is y=x^2, or
f(x) = x^2
so
2f(x) = 2x^2
and everything else, including the scaling, follows from that.
(-1,1) -> (-1,2)
(1,1) -> (1, 2)
(2, 4) -> (2, 8)
See? The y value is doubled, which is kind of obvious when you see that you're going from:
y = x^2
to
y = 2x^2
(And don't worry about being confused; that just means you're paying attention. :-)
But careful: f(-2x) is not -2f(x)!
f(-2x) means might be a shorthand for f(g(x)), where g(x) = -2x.
-2f(x) would be
-2(3/4)x+2(3/2)
=3-(3/2)x
Your parabola is y=x^2, or
f(x) = x^2
so
2f(x) = 2x^2
and everything else, including the scaling, follows from that.
(-1,1) -> (-1,2)
(1,1) -> (1, 2)
(2, 4) -> (2, 8)
See? The y value is doubled, which is kind of obvious when you see that you're going from:
y = x^2
to
y = 2x^2
(And don't worry about being confused; that just means you're paying attention. :-)
Emily
Ohhhh that makes PERFECT sense! :D I'm soo glad I get this now. Thanks so much!! :)