Question

Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

y = x2 – 12x + 12

A. 1 point in common; vertex on x-axis
B. no points in common; vertex above x-axis
C. 2 points in common; vertex below x-axis
D. 2 points in common; vertex above x-axis

Answer 1

@AccessDenied can u help me plz?

Answer 2

@hartnn can u help?

Answer 3

do you know how to find the discrminant of a quadratic expression ?

Answer 4

*discriminant

Answer 5

@hartnn b^2(a*c*4)/b-2 right?

Answer 6

not actually,
for \(ax^2+bx+c=0\)
the discriminant D
\(\large D =b^2-4ac\)

so, first find a,b,c
then calculate D

Answer 7

a= x^2
b= 12x
C= 12

Answer 8

a,b,c are all constants
so,
comparing ax^2 with x^2 gives a =1
bx with -12 x gives b =-12
got this ??

Answer 9

144x-4*x^2*12
4*1*12= 48
144-48=96?

Answer 10

but yes,
\(D = 144-48 = 96\)
is correct :)

Answer 11

How do i put that in a graph tho?

Answer 12

*you don't have to graph it*
so, if D is positive and is NOT a perfect square, then you will have 2 positive roots.

which means it will have 2 points common with x axis

Answer 13

so its D

Answer 14

since your graph opens upwards, and it intersects x axis at 2 points, its vertex will be below x axis

Answer 15

not D, no

Answer 16

Oh ok gotcha so its C then sense its below the axis :)

Answer 17

yes, its C :)

Answer 18

Thanks!! @hartnn