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Graphing a Quadratic

Jaewon Sim
Posted 10 years ago

Hi all, I'm new to Mathematica and I'm not really familiar to the Mathematica functions and the language. Could any of you help me plot a quadratic with the following characteristics:

  • vertex at (1.5, 1.75)
  • roots at roots

('i' means the imaginary number 'i') How would I be able to do this?

Thanks a lot!
POSTED BY: Jaewon Sim
5 Replies
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Dana DeLouis
Dana DeLouis
Posted 10 years ago

But what if you wanted to determine a? y==a(x-b)(x-c)

Depending on one's equation, "sometimes" this may work:

equ =  -24-6 x+3 x^2

FullSimplify[equ,ComplexityFunction->LeafCount]

3 (-4+x) (2+x)

Just to add: Both return the same equation to plot...

MinimalPolynomial[3/2+I Sqrt[7]/2,x]

4-3 x+x^2

MinimalPolynomial[3/2-I Sqrt[7]/2,x]

4-3 x+x^2
POSTED BY: Dana DeLouis
POSTED BY: David Reiss
Jaewon Sim
Jaewon Sim
Posted 10 years ago

Hey, thanks a lot for your help on this! Really helpful advices. What do you think is the best way for me to learn the basics of Mathematica?
POSTED BY: Jaewon Sim

This is your quadradic:

(x - (3/2 + I Sqrt[7]/2)) ((x - (3/2 - I Sqrt[7]/2)))

Expand it to see what it is (and to make sure that it is a real function:

Expand[(x - (3/2 + I Sqrt[7]/2)) ((x - (3/2 - I Sqrt[7]/2)))]

gives

4 - 3 x + x^2

and then plot it:

Plot[4 - 3 x + x^2, {x, -10, 10}]

or you could go to Wolfram|Alpha and type in the equation above (http://www.wolframalpha.com/input/?i=4+-+3+x+%2B+x%5E2)
POSTED BY: David Reiss
Jaewon Sim
Jaewon Sim
Posted 10 years ago

However, if I wanted to make the quadratic equation, wouldn't it be in the form of:

y=a(x-a)(x-b)

In this case, how would I figure out the a value?
POSTED BY: Jaewon Sim
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