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### Course: Algebra (all content)>Unit 9

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.