Answer:
k = 0, k = 4
Step-by-step explanation:
Given equation:
[tex]y=x^2+kx+3k+4x+4[/tex]
Rearrange into standard form [tex]ax^2+bx+c[/tex]:
[tex]\implies y=x^2+kx+4x+3k+4[/tex]
[tex]\implies y=x^2+(k+4)x+(3k+4)[/tex]
Therefore:
[tex]a=1, \quad b=(k+4), \quad c=(3k+4)[/tex]
If the vertex lies in the x-axis, then the quadratic has one (repeating) root at (x, 0). Therefore, we can use the discriminant to find the values of k.
Discriminant
[tex]b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0[/tex]
[tex]\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}[/tex]
[tex]\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}[/tex]
[tex]\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}[/tex]
Therefore, set the discriminant to zero and solve for k:
[tex]\implies (k+4)^2-4(1)(3k+4)=0[/tex]
[tex]\implies k^2+8k+16-12k-16=0[/tex]
[tex]\implies k^2-4k=0[/tex]
[tex]\implies k(k-4)=0[/tex]
[tex]\implies k=0, k=4[/tex]
So the vertex lies in the x-axis when k = 0 or k = 4
