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Graphing quadratics review

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(xh)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.
Incomplete sketch of y=-2(x+5)^2+4
To finish our graph, we need to find another point on the curve.
Let's plug x=4 into the equation.
y=2(4+5)2+4=2(1)2+4=2+4=2
Therefore, another point on the parabola is (4,2).
Final graph of y=-2(x+5)^2+4
Want another example? Check out this video.

Example: Non-vertex form

Graph the function.
g(x)=x2x6
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.
g(x)=x2x60=x2x60=(x3)(x+2)
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.
The average of -2 and 3 is 0.5, which is the x-coordinate of our vertex.
With the x-coordinate figured out, we can solve for y by substituting into our original equation.
g(0.5)=(0.5)2(0.5)6=0.250.56=6.25
Our vertex is at (0.5,6.25), and our final graph looks like this:
Graph of y=x^2-x-6
Want another example? Check out this video.

Practice

Problem 1
Graph the equation.
y=2(x+1)(x1)

Want more practice graphing quadratics? Check out these exercises:

Want to join the conversation?

  • blobby green style avatar for user riley9948
    How even do you guys DO that?!
    (17 votes)
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  • starky seed style avatar for user daily2015
    how to find the vertex
    (10 votes)
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  • leafers ultimate style avatar for user Andrew Escobedo
    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?
    (5 votes)
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  • leaf green style avatar for user Keith Lee
    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?
    (7 votes)
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  • male robot donald style avatar for user Isaiah Walker
    Question? WHere did the word quadratics come from?
    (7 votes)
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  • leaf green style avatar for user just 🐻
    Do i need to learn these, or will it come back to me if i skip parbolas?
    (3 votes)
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  • blobby green style avatar for user Rondina397
    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?
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      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.
      (5 votes)
  • blobby green style avatar for user jalilshair32
    why plugged x=−4x,when you can do y=0
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      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.
      (2 votes)
  • blobby green style avatar for user addjackie
    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?
    (2 votes)
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    • sneak peak blue style avatar for user Shota Oniani
      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:)
      (3 votes)
  • leafers ultimate style avatar for user B  R  I  A  N
    Is there a simpler way to solve this?
    (3 votes)
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    • mr pink green style avatar for user David Severin
      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.
      (0 votes)