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### Course: Algebra 1 (Eureka Math/EngageNY)>Unit 4

Lesson 11: Topic B: Lesson 16: Graphing quadratic equations from the vertex form

Learn how to graph any quadratic function that is given in vertex form. Here, Sal graphs y=-2(x-2)²+5. Created by Sal Khan.

## Want to join the conversation?

• How do you convert a "vertex form" equation into "standard form" equation?
• y = a(x-h)^2 + k is the vertex form equation. Now expand the square and simplify.
You should get y = a(x^2 -2hx + h^2) + k.
Multiply by the coefficient of a and get y = ax^2 -2ahx +ah^2 + k.
This is standard form of a quadratic equation, with the normal a, b and c in ax^2 + bx + c equaling a, -2ah and ah^2 + k, respectively.
• Where did you get the two in the (2,5)?
• It took me 15 minutes to understand this but ,, >>
as you can see , just substitute the value by this
a(x-h)^2 + k
-2(x-2)^2 + 5
Here Sal has taken (h,k) h = 2 and k = 5
don't confuse with the minus sign (-), if there is (-) between x and h thus the value will be positive i.e 2 .If the sign between x and h is positive then the value for 'h' in (h,k) will be negative . The value of the (h,k) is the vertex of parabola .
Don't worry you will get to know it if you analyse for 5 minutes .
You should watch it 3 to 4 times .
• in the graph why did you use 1 and 3 as random numbers?
• You could have used whatever you wanted, but 1 and 3 were convenient as they produced whole numbers when plugged into the equation.
• What if the a value is a fraction? How would I simply graph two coordinates that aren’t the vertex?
• It is not possible to graph a parabola with only 2 points that are not a vertex, unless you also have the average rate of change or both points are on the same side of the parabola. It would be extremely difficult to graph it, and with most points impossible. For Q1, which value are you talking about?
• how would we find the y intercept of an equation in vertex form?
• For any equation, you can find the y-intercept by using x=0 in the equation.
For example: y=2(x+3)^2-4
Set x=0: y=2(0+3)^2-4 = 2(9)-4 = 18-4 = 14
So the y-intercept is at: (0, 14)

Hope this helps.
• We can't make a fully accurate parabola by hand, could we ?
• Well, mathematically speaking, there is no way to draw a "perfect" parabola by hand. Think of it like drawing a "perfect" circle.
• What if A is a fraction? Like; -1/3(x+2)^2 + 3
• Well, the best way to graph this is to first convert this into standard form. We expand the square and combine any like terms:
f(x) = -1/3(x+2)^2 + 3
f(x) = -1/3(x+2)(x+2) + 3
f(x) = (-1/3 x - 2/3)(x+2) + 3
Use some paper and pencil for the next steps, like what I did:
f(x) = -1/3 x^2 - 4/3 x + 5/3
For starters, we can find the vertex first. Let's find the axis of symmetry:
x = (--4/3)/(2(-1/3))
x = -2
Now we plug -2 into the formula, and f(x) = 3. Graph (-2,3), and now, we use the quadratic formula to find the zeroes, which are -5 and 1. We graph the points on the x-axis and connect the three points with a curve. We're done!
• At , how does he determine that the maximum value for y is (2,5)?
Is k always the y term, and is the value of x that makes ax^2 = 0 the y value? I didn't quite get how he came to that.
• y=a(x-h)²+k
I already told you about k which is the vertical shift and (x-h)² which is the horizontal shift on your other question. Now I will tell you about a.

If a is positive, the parabola open up and will have a minimum point at the vertex.
If a is negative, the parabola open down and will have a maximum point at the vertex.
If |a| > 1 , it will have a "skinny" parabola comparing to y=x²
If |a| < 1, it will have a "fat" parabola comparing to y=x²
Try to graph y=x² (|a| =1) and then y=2x² (|a| > 1) and y=(1/2)x² ( |a| < 1) you will what I mean about skinny and fat parabola.
About the vertex, the vertex is determined by (x-h)² and k. The x value that makes x-h=0 will be the x-coordinate of the vertex. K will be the y-coordinate of the vertex.

y=-10(x+4)²-3
This will have a vertex at (-4,-3). a is negative therefore will open down and have a maximum point. |-10| > 1 therefore it's really skinny.

I encourage you try to graph y=a(x-h)²+k with different values it vill help you with a better understanding. You can go here and try your graphs.
https://www.desmos.com/calculator
• Ok. So why does the equation y= -2(x-2)^2+5 have a vertex (h,k) of (2,5) if the value of h is -2?
• That is because the vertex form is:
y = a(x - h)² + k
In this case a = -2
k = 5
h must be 2 because the original form states -h. Therefore, we have:
-h = -2
h = 2
Comment if you have questions.
• What if the first number is a fraction
• It does not change the process if the first number is a fraction.

## Video transcript

We're asked to graph the equation y is equal to negative 2 times x minus 2 squared plus 5. So let me get by scratch pad out so we could think about this. So y is equal to negative 2 times x minus 2 squared plus 5. So one thing, when you see a quadratic or a parabola graph expressed in this way, the thing that might jump out at you is that this term right over here is always going to be positive because it's some quantity squared. Or I should say, it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared. And then we're multiplying it by a negative. So this whole quantity right over here is going to be non positive. It's always going to be less than or equal to 0. So this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. The maximum value for y is 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0. And x minus 2 is equal to 0 when x is equal to 2. So the point 2 comma 5 is the maximum point for this parabola. And it is actually going to be the vertex. So if we were to graph this, so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here is the point 2 comma 5. This is a maximum point, it's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points completely determine a parabola. So that's 1, the vertex, that's interesting. Now, what I'd like to do is just get two points that are equidistant from the vertex. And the easiest way to do that is to maybe figure out what happens when x is equal to 1 and when x is equal to 3. So I could make a table here actually, let me do that. So I care about x being equal to 1, 2, and 3, and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5, so it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1 squared is 1 times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma 3, the point 2 comma 5, and the point 3 comma 3 for this parabola. So let me go back to the exercise and actually put those three points in. And so we have the point 1 comma 3, we have the point 2 comma 5, and we have the point 3 comma 3. And we have now fully determined our parabola.