Quadratic Function[1]

[DanieldeLucena]

Consider the function f(x)=(1/4a)*x^2+x+a, where (a) is a nonzero real number. Choose the alternative which could be the parabola graph of the function.

Correct altervative : Alternative 3 (the blue one)

How I get it ??

 

[Deleted member 4993]

Consider the function f(x)=(1/4a)*x^2+x+a, where (a) is a nonzero real number. Choose the alternative which could be the parabola graph of the function.

Correct altervative : Alternative 3 (the blue one)

How I get it ??

What are the roots of the given equation?

 

[mmm4444bot]

Here's some more hints.

If the vertex of a parabola is the x-intercept, then the two roots are equal.

If the two roots are equal, then the discriminant equals zero.

 

[DanieldeLucena]

But all aternatives have two roots equal ... How do I know the differences between the alternatives?

How I get the conclusion that X > 0 and Y < 0 from the alternative 3 ??
How the parabola has the concave side down, for example ??

I know that delta is equal to zero by

 

[mmm4444bot]

1/4a is not the same as 1/(4a), so I was misled by your original post.

Your work shows that the discriminant (delta) is zero. That's good.

Subhotosh suggested that you find the solution to f(x) = 0. Now would be a good time to do that.

The solution will be an expression with a.

Next, analyze the two cases: a > 0 or a < 0.

For each case, ask yourself, "does the sign of the root place the x-intercept to the left or right of zero? And, does the sign of the leading coefficient indicate that the parabola opens up or down?" Compare with each graph.

You can then see that only one of the three graphs is possible.

If you would like more help, please show your solution to f(x) = 0.

 

[DanieldeLucena]

To X we have[attachment=0:nzruj5ur]CodeCogsEqn(2).gif[/attachment:nzruj5ur]

 

[DanieldeLucena]

But to (a)>0 from (1/4a) we have [attachment=1:3hhyodqm]CodeCogsEqn(3).gif[/attachment:3hhyodqm] So we have ax^2+bx+c with a>0 b>0 c>0
Negative x-vertex [attachment=0:3hhyodqm]CodeCogsEqn(5).gif[/attachment:3hhyodqm]

We don't have aternative with a>0 "parabola opens up" and "negative x-vertex" with c>0

 

[DanieldeLucena]

But to (a)<0 from (1/4a) we have[attachment=1:2ut1qvv8]CodeCogsEqn(4).gif[/attachment:2ut1qvv8]

So we have ax^2+bx+c with a<0 b>0 c<0
Positive x-vertex[attachment=0:2ut1qvv8]CodeCogsEqn(6).gif[/attachment:2ut1qvv8]

So we have the correct alternative with a<0 "parabola opens down" and "positive x-vertex" with c<0

 

[DanieldeLucena]

But I didn't like my explanation ...

Am I right ????

 

[mmm4444bot]

We don't have aternative with a>0 "parabola opens up" and "negative x-vertex"

we have the correct alternative with a<0 "parabola opens down" and "positive x-vertex"

But I [don't] like my explanation

I like it.

Yes, it's correct.

(I don't think that you need to make any statements about b or c.)