I need help with the following question.
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I need help with the following question.

[From: ] [author: ] [Date: 11-09-01] [Hit: ]
-The axis of symmetry is x=-b/(2a).So the vertex is on this line.So the vertex is [-b/2a,B) the discriminant tells you the type of zeroes.If b^2-4ac>0, the graph will cross the x axis twice.......
Consider the general quadratic function f(x)=ax^2+bx+c, with a can’t equal 0.
a. Find the coordinates of the vertex in terms of a, b, and c.
b. Find the conditions on a, b, and c that guarantee that the graph of f crosses the x-axis twice.

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The axis of symmetry is x=-b/(2a). So the vertex is on this line.

Then y= a(-b/2a)^2+b(-b/2a)+c
Y=b^2/(4a)-b^2/2a+c
y=(-b^2)/(4a)+c

Y=(-b^2+4ac)/(4a)
So the vertex is [-b/2a, (-b^2+4ac)/4a]

B) the discriminant tells you the type of zeroes. If b^2-4ac>0, the graph will cross the x axis twice.

( if b^2-4ac<0, then it will not cross the xaxis. If b^2-4ac=0, then it will have one zero.)

Hoping this helps!

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The vertex is located at (-b/(2a) , c - b^2/(4a) ).

b^2 - 4ac > 0 is the condition which guarantees 2 distinct roots.
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