How even do you guys DO that?!

Do some more Khan Academy videos and exercises. You'll get the hang of it sooner of later if you try! Good luck!

how to find the vertex

One way is to complete the square and put in vertex form, Another way is to use -b/2a as the x coordinate and then use that to solve for y.

Question? WHere did the word quadratics come from?

It comes from the Latin word quadratum, which means square.

On paper, how would I find the curve to nicely draw it out? Would I have to plot several points to graph it out by hand or is there another rule?

Yes, the more points you plot, the more accurate you can draw the curve. Some turn the graph paper sideways and draw on one side of the vertex, rotate it 180 degrees and draw on the other side of the vertex.

We only use two point here to graph a parabola but I've heard that it takes three points to define one. Is it true that an infinite number of parabolas can be drawn through just two distinct points? on the coordinate plane?

If you know the vertex and another point, then you also know the reflexive point, so you have 3 points. So two points work as long as one of them is the vertex.

Do i need to learn these, or will it come back to me if i skip parbolas?

Yes, don't skip it. Try to understand it, and it will save you a lot of frustration later on.

On example no.1. How did he came up with x=-4 to plug into the equation? And also in Practise problem no.2, how did he came up with x=0 to plug into the equation?

On Example 1, the vertex was at (-5,4). Sal moved over 1 unit to pick x=-4. I believe he did this to demonstrate the impact of the -2 in front. It reduced the y-value of 4 by 2 units. On practice problem 2, I'm sure he picked x=0 to find the y-intercept of the equation.

why plugged x=−4x,when you can do y=0

Sal has the equation: y = -2(x+5)^2+4. This is vertex form. So, you graph the vertex and then find points to the left and right of the vertex. When Y is isolated already and the equation is in vertex form, it is easier to pick values of X and calculate Y especially when you have a 2nd degree (quadratic) equation or higher degree. The math is usually simpler. If the equation had been in standard form: y=Ax^2+Bx+C, then it is quite common to find the x-intercepts 1st (set y=0 and calculate X). Though, you can also get the vertex pretty easily from this form and work outward from there. The benefit of having the vertex is that you know the highest / lowest point in the graph and you know the graph will be symmetrical as it moves away from the vertex.

I understand how to do this on the quizzes, using the graphing tool, but I'm not sure how I would draw the rest of the parabola "freehand" on paper. How would I know that I extended it perfectly beyond the zero coordinates?

It isn't realistically possible to freehand a parabola, especially on paper. That's why you will not really need to do it that often. In tests and other scenarios where you're asked to graph a parabola the most important things to get right will always be the location(s) of the zero(s) and the vertex. The actual curvature and shape of the parabola is mostly irrelevant if you have those numbers dialed in. If you have all of that information the last bit of info you can use to help draw a more accurate parabola is the "a" coefficient in front of your squared term. That determines how "squished" or "stretched" your parabola actually is. You can visualize this by going to an online graphing calculator (or other alternative) and playing around with that value. Hope this helps:)

Main content

Graphing quadratics review

Google Classroom

The graph of a quadratic function is a parabola, which is a "u"-shaped curve. In this article, we review how to graph quadratic functions.

The graph of a quadratic function is a parabola, which is a "u"-shaped curve:

In this article, we review how to graph quadratic functions.

Looking for an introduction to parabolas? Check out this video.

Example 1: Vertex form

Graph the equation.

y = - 2 (x + 5)^{2} + 4

This equation is in vertex form.

y = a (x - h)^{2} + k

This form reveals the vertex,

(h, k)

, which in our case is

(- 5, 4)

It also reveals whether the parabola opens up or down. Since

a = - 2

, the parabola opens downward.

This is enough to start sketching the graph.

To finish our graph, we need to find another point on the curve.

Let's plug

x = - 4

into the equation.

\begin{aligned} y & = - 2 (- 4 + 5)^{2} + 4 \\ = - 2 (1)^{2} + 4 \\ = - 2 + 4 \\ = 2 \end{aligned}

Therefore, another point on the parabola is

(- 4, 2)

Want another example? Check out this video.

Example: Non-vertex form

Graph the function.

g (x) = x^{2} - x - 6

First, let's find the zeros of the function—that is, let's figure out where this graph

y = g (x)

intersects the

x

-axis.

\begin{aligned} g (x) & = x^{2} - x - 6 \\ 0 & = x^{2} - x - 6 \\ 0 & = (x - 3) (x + 2) \end{aligned}

So our solutions are

x = 3

and

x = - 2

, which means the points

(- 2, 0)

and

(3, 0)

are where the parabola intersects the

x

-axis.

To draw the rest of the parabola, it would help to find the vertex.

Parabolas are symmetric, so we can find the

x

-coordinate of the vertex by averaging the

x

-intercepts.

With the

x

-coordinate figured out, we can solve for

y

by substituting into our original equation.

\begin{aligned} g (0.5) & = (0.5)^{2} - (0.5) - 6 \\ = 0.25 - 0.5 - 6 \\ = - 6.25 \end{aligned}

Our vertex is at

(0.5, - 6.25)

, and our final graph looks like this:

Want another example? Check out this video.

Practice

Problem 1

Graph the equation.

y = 2 (x + 1) (x - 1)

Want more practice graphing quadratics? Check out these exercises:

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riley9948
Posted 4 years ago. Direct link to riley9948's post “How even do you guys DO t...”
How even do you guys DO that?!
Button navigates to signup pageComment on riley9948's post “How even do you guys DO t...”
(16 votes)
Answer
- CreatorOfBob
  Posted 4 years ago. Direct link to CreatorOfBob's post “Do some more Khan Academy...”
  Do some more Khan Academy videos and exercises. You'll get the hang of it sooner of later if you try! Good luck!
  Comment on CreatorOfBob's post “Do some more Khan Academy...”
  (44 votes)
daily2015
Posted 7 years ago. Direct link to daily2015's post “how to find the vertex”
how to find the vertex
Button navigates to signup pageComment on daily2015's post “how to find the vertex”
(9 votes)
Answer
- David Severin
  Posted 7 years ago. Direct link to David Severin's post “One way is to complete th...”
  One way is to complete the square and put in vertex form, Another way is to use -b/2a as the x coordinate and then use that to solve for y.
  Comment on David Severin's post “One way is to complete th...”
  (20 votes)
Andrew Escobedo
Posted 5 years ago. Direct link to Andrew Escobedo's post “We only use two point her...”
We only use two point here to graph a parabola but I've heard that it takes three points to define one. Is it true that an infinite number of parabolas can be drawn through just two distinct points? on the coordinate plane?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “If you know the vertex an...”
  If you know the vertex and another point, then you also know the reflexive point, so you have 3 points. So two points work as long as one of them is the vertex.
  Comment on David Severin's post “If you know the vertex an...”
  (13 votes)
Isaiah Walker
Posted 5 years ago. Direct link to Isaiah Walker's post “Question? WHere did the w...”
Question? WHere did the word quadratics come from?
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(7 votes)
Answer
- Spencer York
  Posted 2 years ago. Direct link to Spencer York's post “It comes from the Latin w...”
  It comes from the Latin word quadratum, which means square.
  Button navigates to signup page
  (4 votes)
Keith Lee
Posted 5 years ago. Direct link to Keith Lee's post “On paper, how would I fin...”
On paper, how would I find the curve to nicely draw it out? Would I have to plot several points to graph it out by hand or is there another rule?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “Yes, the more points you ...”
  Yes, the more points you plot, the more accurate you can draw the curve. Some turn the graph paper sideways and draw on one side of the vertex, rotate it 180 degrees and draw on the other side of the vertex.
  Button navigates to signup page
  (4 votes)
just 🐻
Posted 2 years ago. Direct link to just 🐻's post “Do i need to learn these,...”
Do i need to learn these, or will it come back to me if i skip parbolas?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Spencer York
  Posted 2 years ago. Direct link to Spencer York's post “Yes, don't skip it. Try t...”
  Yes, don't skip it. Try to understand it, and it will save you a lot of frustration later on.
  Button navigates to signup page
  (7 votes)
Rondina397
Posted 4 years ago. Direct link to Rondina397's post “On example no.1. How did ...”
On example no.1. How did he came up with x=-4 to plug into the equation?

And also in Practise problem no.2, how did he came up with x=0 to plug into the equation?
Button navigates to signup pageComment on Rondina397's post “On example no.1. How did ...”
(3 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “On Example 1, the vertex ...”
  On Example 1, the vertex was at (-5,4). Sal moved over 1 unit to pick x=-4. I believe he did this to demonstrate the impact of the -2 in front. It reduced the y-value of 4 by 2 units.
  
  On practice problem 2, I'm sure he picked x=0 to find the y-intercept of the equation.
  Button navigates to signup page
  (5 votes)
jalilshair32
Posted a year ago. Direct link to jalilshair32's post “why plugged x=−4x,when yo...”
why plugged x=−4x,when you can do y=0
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “Sal has the equation: y =...”
  Sal has the equation: y = -2(x+5)^2+4. This is vertex form. So, you graph the vertex and then find points to the left and right of the vertex.
  
  When Y is isolated already and the equation is in vertex form, it is easier to pick values of X and calculate Y especially when you have a 2nd degree (quadratic) equation or higher degree. The math is usually simpler.
  
  If the equation had been in standard form: y=Ax^2+Bx+C, then it is quite common to find the x-intercepts 1st (set y=0 and calculate X). Though, you can also get the vertex pretty easily from this form and work outward from there.
  
  The benefit of having the vertex is that you know the highest / lowest point in the graph and you know the graph will be symmetrical as it moves away from the vertex.
  Button navigates to signup page
  (2 votes)
addjackie
Posted a year ago. Direct link to addjackie's post “I understand how to do th...”
I understand how to do this on the quizzes, using the graphing tool, but I'm not sure how I would draw the rest of the parabola "freehand" on paper. How would I know that I extended it perfectly beyond the zero coordinates?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Shota Oniani
  Posted a year ago. Direct link to Shota Oniani's post “It isn't realistically po...”
  It isn't realistically possible to freehand a parabola, especially on paper. That's why you will not really need to do it that often. In tests and other scenarios where you're asked to graph a parabola the most important things to get right will always be the location(s) of the zero(s) and the vertex. The actual curvature and shape of the parabola is mostly irrelevant if you have those numbers dialed in.
  
  If you have all of that information the last bit of info you can use to help draw a more accurate parabola is the "a" coefficient in front of your squared term. That determines how "squished" or "stretched" your parabola actually is. You can visualize this by going to an online graphing calculator (or other alternative) and playing around with that value.
  
  Hope this helps:)
  Comment on Shota Oniani's post “It isn't realistically po...”
  (3 votes)
B R I A N
Posted 5 years ago. Direct link to B R I A N's post “Is there a simpler way to...”
Is there a simpler way to solve this?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “Are you talking about the...”
  Are you talking about the vertex form or the standard form? Learning the parent function helps graph vertex form by using the idea of scale factor. So parent function has (0.0)(1,1) and (-1,1), (2,4) and (-2,4) because 2^2=(-2)^2=4. With a scale factor of -2, multiply the y times 2 and "act like the vertex" is the origin. With the vertex at (-5,4) and a=-2, go down 2 over 1 both left and right from vertex (1,1) and (-1,1) switches to (1,-2) and (-1,-2) from vertex, you might be able to go down 8 over 2 both left and right for more points (2,4) switches to (2,-8).
  For the standard form, you could complete the square to get into vertex form.
  Comment on David Severin's post “Are you talking about the...”
  (0 votes)

Integrated math 2

Course: Integrated math 2 > Unit 3

Example 1: Vertex form

Example: Non-vertex form

Practice

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