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## Get ready for Algebra 2

### Course: Get ready for Algebra 2 > Unit 3

Lesson 11: Features and forms of quadratic functions- Forms & features of quadratic functions
- Worked examples: Forms & features of quadratic functions
- Features of quadratic functions: strategy
- Vertex & axis of symmetry of a parabola
- Finding features of quadratic functions
- Features of quadratic functions
- Graph parabolas in all forms
- Graphing quadratics review

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# Graphing quadratics review

CCSS.Math: , , , ,

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.**

This equation is in vertex form.

This form reveals the vertex, left parenthesis, start color #11accd, h, end color #11accd, comma, start color #1fab54, k, end color #1fab54, right parenthesis, which in our case is left parenthesis, minus, 5, comma, 4, right parenthesis.

It also reveals whether the parabola opens up or down. Since start color #e07d10, a, end color #e07d10, equals, minus, 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, equals, minus, 4 into the equation.

Therefore, another point on the parabola is left parenthesis, minus, 4, comma, 2, right parenthesis.

*Want another example? Check out this video.*

### Example: Non-vertex form

**Graph the function.**

First, let's find the zeros of the function—that is, let's figure out where this graph y, equals, g, left parenthesis, x, right parenthesis intersects the x-axis.

So our solutions are x, equals, 3 and x, equals, minus, 2, which means the points left parenthesis, minus, 2, comma, 0, right parenthesis and left parenthesis, 3, comma, 0, right parenthesis 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.

Our vertex is at left parenthesis, 0, point, 5, comma, minus, 6, point, 25, right parenthesis, and our final graph looks like this:

*Want another example? Check out this video.*

## Practice

*Want more practice graphing quadratics? Check out these exercises:*

## Want to join the conversation?

- How even do you guys DO that?!(8 votes)
- Do some more Khan Academy videos and exercises. You'll get the hang of it sooner of later if you try! Good luck!(19 votes)

- how to find the vertex(8 votes)
- 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.(16 votes)

- 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)
- 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.(8 votes)

- 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?(4 votes)
- 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.(2 votes)

- Question? WHere did the word quadratics come from?(4 votes)
- It comes from the Latin word quadratum, which means square.(1 vote)

- 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?(2 votes)- 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)

- what do you do with the excess numbers(2 votes)
- y=−2(x−1)^2−4 if we try to find roots it will give no real solution we can only graph using vertex form??(1 vote)
- You could graph from standard form also, the x value of the vertex is -b/2a (recognize this from quadratic formula?), then substitute that in to find y value of vertex, find additional points from there just like with the vertex form. You are correct because the -2 indicates it opens downward and the vertex is at (1,-4).(2 votes)

- how to find the vertex of y=2/3x(x-6)(1 vote)
- Since the solutions would be at 0 and 6, these are symmetric points, so the x value must be 1/2 way in between these two which is at x=3. Then if x=3, the 2/3(3)(3-6) gives 2(-3) = -6, so the vertex is at (3,-6).

Or if you multiply it out, you get 2/3x^2 - 4x, factor out 2/3 to get 2/3(x - 6x) note that 2/3*6=4. Complete the squaure by dividing 6 by 2 to get 3 and square it to get 9. So 2/3(x - 6x + 9 - 9) but 2/3(-9)=-6. Thus, vertex form is (x-3)^2 - 6 which again gives (3,-6).(2 votes)

- how do i find the x intercepts in vertex form(1 vote)
- x intercepts are the 0s or the roots. So set the equation equal to 0 and solve. keep in mind if you take the square root of a number you get a positive and negative result. sqrt(4) = positive and negative 4(1 vote)