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## Get ready for Algebra 2

### Unit 3: Lesson 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?

• 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.
• 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 .
• What if the first number is a fraction
• It does not change the process if the first number is a fraction.
• How do you find A in the vertex form equation when you are only given the vertex point and a point on the parabola?
• Good question. You will have an equation of the form:
𝑦 = 𝑎(𝑥 - ℎ)² + 𝑘
You are given (ℎ, 𝑘) and an coordinate (𝑥, 𝑦). Notice, if you plug in all these values into the equation, you'll be left with only one variable which is 𝑎. You can therefore algebraically manipulate the equation and solve for 𝑎.
• Is y basically f(x)?
• Yes. When an equation is written in function notation, the "y" variable is replaced with f(x). So, f(x) = y.
• 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.
• I still need a little help finding the minimum values of a parabola when the equation is in vertex form.
(1 vote)
• So when Sal finds the vertex at (2,5) at that is the maximum point. If we flip that parabola into an upward-opening parabola the vertex would be at (-2,-5) so that would be the minimum point.
(1 vote)
• how would we find the y intercept of h(x)=0.5(x+4)^2 +1 ?