Question

Teaching how to graph quadratics. What is the simplest way to explain how to graph a quadratic equation?

Answer 1

quadratic equation usually means
\[ax^2+bx+c=0\] quadratic function (or graph)
\[y=ax^2+bx+c\]
1) open up or open down \(a>0,a<0\)
2) vertex \((-\frac{b}{2a},f(-\frac{b}{2a})\)
3) y intercept \((0,c)\)
4) x intercepts (if they exist), \(0,\frac{-b\pm\sqrt{b^2-4ac}}{2})\)
5) plot a point or two

Answer 2

u a teacher?

Answer 3

I like to put it in the form
\(y=a(x-h)^2+k\)
and then connect this with the \(y=mx+b\) form of a line.

Answer 4

Satellite, yes, I know all of that, but it's not a very straightforward way to teach it, I don't think.
Quark, yes I am a teacher.
Markriggs, I like to use vertex form too, and I have little trouble teaching them to graph equations already in vertex form. I don't find it very simple to explain to them how to put the graph in vertex form to start with, however.

I'm wondering if it wouldn't be best to just teach them to plot the vertex at (-b/2a) and then plot a few points out from there.

Answer 5

It is a little difficult at first, but the same skills are used for putting a circle in standard form.

Answer 6

i would say the simplest way to teach graphing a quadratic is to have a firm foundation for the graphing linear equation since all that changes is a bend.

Answer 7

Personally the way they taught me was to plot the vertex (of course use -b/2a) and then create a T-chart using two values to the right and two to the left of the x-value. Substitute those values for x and find the y values of each. so then I'd have the vertex and 4 other points.
Though I honestly believe using the quadratic equation is pretty straightforward and makes finding the solutions a lot easier. Amazingly it was really simple to remember.

I just learned about quadratic equations several weeks ago so its fresh in my mind.

Answer 8

Jazy, that's the method I'm leaning towards myself.
The quadratic formula is fine, and it will give you the x-intercepts, but it's not always easy to see the shape of the graph from those points. =/

Answer 9

When I think of having to graph a quadratic equation I look at the form:
\[ax^2 + bx + c\]
I know that if
a = positive: the parabola opens upward U and the vertex is at a minimum point.
a = negative: the parabola opens downward and the vertex is at a maximum point.
Once I find the vertex and other points, I already have an idea what the graph will look like.
Personally, remembering those basics made it easier for me to understand.

Answer 10

I'm not too experienced in teaching this stuff lol. I just finished Algebra 1 and found it pretty simple to learn this without giving it much thought. Of course i don't know what it is your teaching. If its just the basic Algebra lessons on graphing Quadratic equations, I think your students will understand if they know what a graph looks like based on the equation, and if they know how to find other points.

It IS kind of hard to explain lol:)

Answer 11

Yes, well, I think you have a good perspective as a student who has recently learned it.

Answer 12

I took Algebra on an online course(which tends to be more difficult to understand math) just by reading and seeing step to step instructions. They only tell you the keys to quadratic equations. They don't go very much into detail, yet I passed with a high grade.

Answer 13

You seem like an enthusiastic teacher -unlike other unpleasant ones who don't teach anything- so I think one way or another your students will get what your teaching:)

Answer 14

I will tell them a little bit about "parabola" first.

Answer 15

vertex form is a pain in the retriceif the problem is given as
\[y=ax^2+bx+c\] because it entails completing the square and keeping track to get \(k\)
easier to say that if you have
\[y=ax^2+bx+c\] then vertex form is going to be
\[y=a(x-h)^2+k\] where
\[h=-\frac{b}{2a}\] and
\[k=f(h)\] since this is usually much easier to compute

Answer 16

Yeah, I think I've come to the same conclusion. I don't usually like just giving them a "formula" without explaining why it works. But I think this will make some sense to them and they'll have more success with it than with trying to put into vertex form.

Answer 17

if you want to motivate why the vertex is what it is start with
\[y=x^2\] then go to
\[y=(x-2)^2\] then go to
\[y=(x-2)^2+3\]and it will be clearer.

Answer 18

try something like the attached sheet
it is not about graphing, but it will help with the vertex

Answer 19

Oh, wow. I like that a lot. Very logical explanation of why vertex form works. =D

Answer 20

glad you like it. it could use some improvement, but it might do as a start

Answer 21

in particular you could incorporate graphing. if you use something similar in class, you might point out what a colossal pain it is to keep track of the \(k\) term when completing the square, (after having done an annoying example) then write say
\[y=2x^2+6x-1\]
\[y=2(x^2+3x)-1\]
\[y=2(x+\frac{3}{2})^2+k\] and then ask if there an easier way to find k
now that you know the value of x that will get the minimum
i can almost guarantee that someone paying attention will say "plug it in" and inf not, you can prompt by asking "if i know the first coordinate, how do i find the second?"