Answer:
y = 2x^2 + 8x + 8
Step-by-step explanation:
The graph touches the x axis at only one point.
so there is only one real solution.
If there is only one real solution then determinant =0
Now we find out the equation that has determinant 0
Determinant is [tex]b^2 - 4ac[/tex]
Let find b^2 - 4ac for each equation
(a) [tex]y = 9x^2 + 6x + 4[/tex]
a= 9 , b = 6 and c=4
[tex]b^2-4ac= 6^2 - 4(9)(4) = -108[/tex]
determinant not equal to 0
(b) [tex]y = 6x^2 – 12x – 6[/tex]
a= 6 , b = -12 and c=-6
[tex]b^2-4ac= (-12)^2 - 4(6)(-6) = 288[/tex]
determinant not equal to 0
(c) [tex]y = 3x^2 + 7x + 5[/tex]
a= 3 , b = 7 and c=5
[tex]b^2-4ac= (7)^2 - 4(3)(5) = -11[/tex]
determinant not equal to 0
(d) [tex]y = 2x^2 + 8x + 8[/tex]
a= 2 , b = 8 and c=8
[tex]b^2-4ac= (8)^2 - 4(2)(8) = 0[/tex]
determinant equal to 0. So there is only one real solution.
