Find Quadratic Polynomial from Three Given Points

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happysmiler07

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Suppose a parabola has x-intercepts (-2,0) and (4,0) and y-intercept (0,3). How would you find the formula for the parabola?
 
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happysmiler07 said:
Suppose a parabola has x-intercepts (-2,0) and (4,0) and y-intercept (0,3). How would you find the formula for the parabola?
Click to expand...
Plug each of the three points into y = ax^2 + bx + c. This will give you a system of equations in a, b, and c. Solve for the values of these coefficients. This gives you the formula. ;)
 
Denis said:
I probably need another coffee....BUT:
GIVEN: x-intercepts (-2,0) and (4,0) ; so:
(x+2)(x-4) = 0
x^2 - 2x - 8 = 0

y intercept = -8 ; ye olde upward-opener-parabola...

What did I miss...how long in the corner?
Click to expand...

y = -3/8*(x+2)*(x-4)

y_intercept = (0,3)

to the corner for 3/8 minutes .......
 
happysmiler07 said:

Can you explain a little more on how you use the y-intercept to find the constant -3/8?
Click to expand...

It's not easy to explain, when you have not shown any work. What have you done so far?


Did you copy the equation from Denis' first reply, without understanding what it means?

It means a(x+2)(x-4) = 0


Did you copy the answer in Subhotosh's reply, but you don't understand what -3/8 means?

Stapel already answered your questions, by giving you some steps to finish the exercise.

Denis suggested a different approach.

Which method are you using?

:idea: If you follow the forum guidelines, these questions won't be an issue. Here's a link to the summary page.

Cheers
 
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