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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

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:
A coordinate plane. The x- and y-axes both scale by one. The graph is the function x squared. The function is a parabola that opens up. The function decreases through negative two, four and negative one, one. The vertex of the function is plotted at zero, zero, then the function increases through one, one and two, four.
Looking for an introduction to parabolas? Check out this video.

### Example 1: Vertex form

Graph the equation.
y, equals, minus, 2, left parenthesis, x, plus, 5, right parenthesis, squared, plus, 4

This equation is in vertex form.
y, equals, start color #e07d10, a, end color #e07d10, left parenthesis, x, minus, start color #11accd, h, end color #11accd, right parenthesis, squared, plus, start color #1fab54, k, end color #1fab54
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.
A coordinate plane. The x- and y-axes both scale by one. The graph is the function negative two times the sum of x plus five squared plus four. The function is a parabola that opens down. The vertex of the function is plotted at the point negative three, four and there are small lines leaving toward the rest of the function.
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, equals, minus, 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 left parenthesis, minus, 4, comma, 2, right parenthesis.
A coordinate plane. The x- and y-axes both scale by one. The graph is the function negative two times the sum of x plus five squared plus four. The function is a parabola that opens down. The vertex of the function is plotted at the point negative three, four. Another point is plotted at negative four, two.
Final graph of y=-2(x+5)^2+4
Want another example? Check out this video.

### Example: Non-vertex form

Graph the function.
g, left parenthesis, x, right parenthesis, equals, x, squared, minus, x, minus, 6

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.
\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, 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.
A coordinate plane. The x- and y-axes both scale by one. The points negative two, zero and three, zero are plotted.
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.
A coordinate plane. The x- and y-axes both scale by one. The points negative two, zero and three, zero are plotted. A point in plotted in the middle of these point at zero point five, zero.
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.
\begin{aligned} g(\blueD{0.5})&=(\blueD{0.5})^2-(\blueD{0.5})-6 \\\\ &=0.25-0.5-6 \\\\ &=-6.25 \end{aligned}
Our vertex is at left parenthesis, 0, point, 5, comma, minus, 6, point, 25, right parenthesis, and our final graph looks like this:
A coordinate plane. The x- and y-axes both scale by one. The graph is the function x squared minus x minus six. The function is a parabola that opens up. The vertex of the function is plotted at the point zero point five, negative six point two-five. The x-intercepts are also plotted at negative two, zero and three, zero.
Graph of y=x^2-x-6
Want another example? Check out this video.

## Practice

Problem 1
• Current
Graph the equation.
y, equals, 2, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 1, right parenthesis

Want more practice graphing quadratics? Check out these exercises:

## Want to join the conversation?

• 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.
• 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.
• 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.
• Question? WHere did the word quadratics come from?
• 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?
• 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.
• what do you do with the excess numbers