Question

Pleaseee help????😭

State the conditions on the values a and k so that the parabola y=a(x-h)^2+k opens downward and intersects the x-axis in two points

Answer 1

a must be negative so that the parabola slopes downward and k must be greater than 0 so that the parabola intersects the x axis.
\[a<0 & k>0\]

Answer 2

please, note that,the condition on a, is:
\[a<0\]
as @Jaguar1998 well wrote,
so if I set:
\[a=-\alpha ^{2}\]
I can re-write your parabola as follows:
\[y=-\alpha ^{2}(x-h)^{2}+k\]
and it intersects the x-axis, if and only if:
\[\alpha ^{2}(x-h)^{2}=k\]
from which:
\[x=h \pm \frac{ \sqrt{k} }{ \alpha } \]
and as you can see, the solution exist if \[k \ge 0\]

Answer 3

oops...the solutions can exist if...

Answer 4

Michel your answer is flawed in many ways. for one, the negative next to alpha squared must also be squared. Also if k is equal to zero then the graph only intersects the x axis at one point.

Answer 5

for your first question, I think you are wrong, because I have to solve this equation:
\[-\alpha ^{2}(x-h)^{2}+k=0\]
for your second question you are wrong, because the text ofquestion of @jenniferjuice doesn't specify if the points have to be distinct, so if k=0, we have to solve this equation:

Answer 6

sorry my keyboard was being stupid im on my laptop now

Answer 7

sorry, for your second question you are right! k>0 only @Jaguar1998

Answer 8

okay so a and k are 0?

Answer 9

Michele, the author specified that the graph must intersect the x axis at two points. Also -alpha^2 does not equal a because you are changing the signs. (-alpha)^2 can equal a. but the negative sign must be in the parenthesis

Answer 10

a<0 and k>0

Answer 11

my original answer :)

Answer 12

yes! youy are right! @Jaguar1998
if k=0, our parabola degenerates in a couple of lines!

Answer 13

please I wrote \[a=-\alpha ^{2}\]
in order to work with a positive quantity
@Jaguar1998

Answer 14

...or in order to explicit the negative sign