18,331 views
24 votes
24 votes
Please help I will mark branlist

Please help I will mark branlist-example-1
User Knasterbax
by
3.1k points

2 Answers

20 votes
20 votes

Answer:

False. It has no solutions.

Step-by-step explanation:

Hope it helps! =D

User Piyush Dhamecha
by
3.1k points
16 votes
16 votes

Answer:If a graph of a quadratic,

f

(

x

)

, does not have an x-intercept

then

f

(

x

)

=

0

has no Real solutions.

Step-by-step explanation:

The x-axis is composed of all points for which

f

(

x

)

(or, if you prefer,

y

) is equal to

0

If the graph of

f

(

x

)

does not have an x-intercept

then it has no (Real) points for which

f

(

x

)

=

0

Step-by-step explanation:

No x-intercept means that it does not cross the x-axis. Thus two solutions is definitely ruled out.

However, if you DO NOT INCLUDE a point of coincidence (Vertex coincides with the x-axis) in the phrase "does not have an x-intercept". Then there could be a single value solution if you equate the quadratic to 0. Some people say that it still has two in such a case but they are both the same value. I do not like this way of thinking!

On the other hand, if you DO INCLUDE a point of coincidence in the phrase, then the plot does not cross the x-axis nor does any point on the curve coincide with it. In such an interpretation there is NO SOLUTION THAT IS REAL. hopes this helps

User Toyota Supra
by
2.6k points