Can we use the "discriminant way" for this...?

Question

Can we use the "discriminant way" for this...?
The curve y = (k-6)x^2 - 8x + k cuts the x-axis at two points and has a minimum point. Find the range of values of k.

Answer 1

Yes...

a should be greater then 0

k-6=a
k-6 greater than 0
k greater than 6

b^2-4ac greater than or equal to 0
(-8)^2-4(k-6)(k) greater than or equal to 0
64-4k^2+24k greater than or equal to 0

16-k^2+6k greater than or equal to 0
k^2-6k-16 less than or equal to 0

equate to zero:
(k-8)(k+2) = 0
critical points: k=8, k=-2

number line:
substitute a value from each of the segments to (k-8)(k+2):

.+...........-..............+
-----(-2)---------(8)------


*product should be negative so we get the values in between:

-2 and 8 including -2 and 8 but since the condition should be greater than 6, then k would be:

(6,8]

Answer 2

Yes we can

when x axis we have

y =0
so (k-6)x^2-8x +k =0

for this to have 2 solutions

discrminant(comapring with ax^2+bx+c , b^2-4ac)
= 8^2-4k(k-6) > 0
or 64-4k^2+24k >0
or 16-k^2+6k>0
or 16-(k^2+6k+9)+9 >0
or 25-(k+3)^2 >0
or (k+3)^2 < 25

or -5 < k+3 < 5
-8 < k < 2
for these values it cuts x asis at 2 points
if k = -8 or 2 it is tangent and we can find wheter minimum or maximum
if k is outsde range no cut

Answer 3

That's the easiest. Calculate b² - 4ac (in standard form) and then find out the range of k for which it's > 0.

8²-4(k-6)k = 64-4k²+24k = -4k²+24k+64 (in standard form). Now find k values for which this function is > 0
(Hint: Since the coefficient of the k² term is negative, the parabola opens downwards and the k values needed will be between the roots ☺)


Doug

Answer 4

discriminant must be greater than zero

the arms are up

k-6>0

Answer 5

Do your own homework cheat.