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