Graphs of Functions (quadratic)

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jpanknin

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I'm having trouble understanding what effect the middle term ([MATH]2x[/MATH]) in the function below has on the graph:

[MATH]f(x) = x^2 + 2x + 1[/MATH]
I'm trying to intuitively understand. I have a good grasp on the [MATH]x^2[/MATH] graph and also the [MATH]+1[/MATH] shift and can draw the graph [MATH]f(x) = x^2 + 1[/MATH] just by looking at the equation, but I can never intuitively grasp where the graph will end up when the middle term is included. Any help in understanding is much appreciated.

Note: I know what the graph looks like and I'm not just interested in this specific equation, but more the intuition behind the shift the [MATH]2x[/MATH]/middle term causes.
 
jpanknin said:
I'm having trouble understanding what effect the middle term ([MATH]2x[/MATH]) in the function below has on the graph:

[MATH]f(x) = x^2 + 2x + 1[/MATH]
I'm trying to intuitively understand. I have a good grasp on the [MATH]x^2[/MATH] graph and also the [MATH]+1[/MATH] shift and can draw the graph [MATH]f(x) = x^2 + 1[/MATH] just by looking at the equation, but I can never intuitively grasp where the graph will end up when the middle term is included. Any help in understanding is much appreciated.

Note: I know what the graph looks like and I'm not just interested in this specific equation, but more the intuition behind the shift the [MATH]2x[/MATH]/middle term causes.
Click to expand...
I don't generally try to think about the effect of the middle term by itself, but it is interesting to ponder.

I would think in terms of completing the square. Given the original function [MATH]f(x) = x^2[/MATH], and the function [MATH]g(x) = x^2 + 1[/MATH] which is shifted up by 1, the function [MATH]h(x) = x^2 + 2x + 1 = (x+1)^2[/MATH] is shifted left 1 compared to f, and therefore left 1 and down 1 compared to g. In general, adding a middle term has that sort of compound effect on the graph.

That compound effect (shifting diagonally, and in an amount that depends on the other coefficients) is why I don't think about it. I just rewrite the function by completing the square, as I did here, and look at a, h, k in [MATH]a(x-h)^2+k[/MATH] rather than a, b, c in [MATH]ax^2+bx+c[/MATH].
 
Graph many of these parabola and become the expert on this question. If you investigate this yourself in the end you would learn more than you can learn from us.
 
Dr.Peterson said:
I don't generally try to think about the effect of the middle term by itself, but it is interesting to ponder.

I would think in terms of completing the square. Given the original function [MATH]f(x) = x^2[/MATH], and the function [MATH]g(x) = x^2 + 1[/MATH] which is shifted up by 1, the function [MATH]h(x) = x^2 + 2x + 1 = (x+1)^2[/MATH] is shifted left 1 compared to f, and therefore left 1 and down 1 compared to g. In general, adding a middle term has that sort of compound effect on the graph.

That compound effect (shifting diagonally, and in an amount that depends on the other coefficients) is why I don't think about it. I just rewrite the function by completing the square, as I did here, and look at a, h, k in [MATH]a(x-h)^2+k[/MATH] rather than a, b, c in [MATH]ax^2+bx+c[/MATH].
Click to expand...
Thank you. The diagonal shift is what confuses me (not for this graph, but for other similar equations with middle terms - I chose a simple one for illustration purposes). For example, [MATH]f(x) = x^2 + 8x + 1[/MATH] shifts down and left. I just can't wrap my mind around WHY that shift occurs. All the material I'm studying goes over the basic transformations, but no example has discussed the intuition around this issue.
 
We know what the constant does, it shift the graph up by whatever the constant is. So ignore functions with constants.
So we are left with functions of the form ax^2 + bx = a(x^2 +b/a). We know, or should know, what the a in front does. This leaves us with functions of the form y = x^2 + bx. Look closely at what b does!
y= x^2 + bx = x(x+b) = 0 when x = 0 or x=b.
Assume b>0, then what happens if b is replaced with a larger value? With a value between 0 and b? less than b? What if b<0.

Note that if y = x^2 + bx + c, then whatever your answers are for my question above it will now be true when y=c
 
jpanknin said:
Thank you. The diagonal shift is what confuses me (not for this graph, but for other similar equations with middle terms - I chose a simple one for illustration purposes). For example, [MATH]f(x) = x^2 + 8x + 1[/MATH] shifts down and left. I just can't wrap my mind around WHY that shift occurs. All the material I'm studying goes over the basic transformations, but no example has discussed the intuition around this issue.
Click to expand...
The shift occurs in order to move the vertex (think about the formula for the vertex, and the quadratic formula) without changing either the y-intercept (determined by c) or the shape (determined by b). Try changing b in this graph while keeping a and c at whatever fixed values you wish, and think about it:

www.desmos.com

FMH126567

www.desmos.com www.desmos.com

As I implied, I don't try to wrap my mind around exactly what it does; instead, I put the function in a form whose parameters are all directly meaningful.
 
jpanknin said:
Thank you. The diagonal shift is what confuses me (not for this graph, but for other similar equations with middle terms - I chose a simple one for illustration purposes). For example, [MATH]f(x) = x^2 + 8x + 1[/MATH] shifts down and left. I just can't wrap my mind around WHY that shift occurs. All the material I'm studying goes over the basic transformations, but no example has discussed the intuition around this issue.
Click to expand...
Did you look at the link that I supplied in reply #2? Because I am an advocate of discovery learning, I hoped that you might use the WolfranAlpha site to explore the various plots of \(x^2+bx+c\). Your post have been about what effect the term \(bx\) has to do with the function.
If you use the link to explore \(x^2+bx+c\) where \(c=8~\&~b=1,2,-4,-7\) In every one of those cases the graph passes through the point \((0,8)\) Why is that the case?
 
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