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#1 2005-12-13 11:22:49
- Anastasia
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quadratics
2x^2+ 4x + 1 = 0
I need help finding the y and x intercepts and vertex!
I also need to use the quadratic formula to find, whatever that formula finds.... Please help!
#2 2005-12-13 11:33:49
- Chemist
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Re: quadratics
you mean finding the roots of this quad eq ?
General formula :
ax^2 + bx + c = 0
First find Delta Δ = b^2 - 4ac
x1 = ( -b + √Δ ) / 2a
x2 = ( -b - √Δ ) / 2a
If b is even , you can take delta = b^2 - ac , if i remember correctly, and instead of 2a, u use a in the denominator.
Anyway the roots should be :
x1 = - 0.29
x2 = -1.7
Concerning the intercepts, just equate x and y to 0 for a general equation. In this case, this is a parabola, so the roots found are x intercept and the y intercept is 0.
Last edited by Chemist (2005-12-13 11:55:37)
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
#3 2005-12-13 23:06:51
- mathsyperson
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Re: quadratics
The x intercepts are as Chemist already said, but the y intercept is 1, because the equation is 2x² + 4x +1.
I'm not sure what you mean by vertex, but I'll guess you mean the turning point. To find that, you need to complete the square.
Completing the square for this example would give 2(x + 1)² + 1. (x + 1)² is always positive, so it is smallest when it is equal to 0. This is when x = -1 and so that is your turning point.
Why did the vector cross the road?
It wanted to be normal.
#5 2005-12-14 01:53:18
- mathsyperson
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Re: quadratics
Yes, but what is the line intersecting with? We've already calculated where it crosses the axes. I agree that vertex wouldn't normally mean 'minimum point', but I couldn't think of what else it could mean.
Why did the vector cross the road?
It wanted to be normal.