Question

Please solve with explanation high points

Answer 1

k = 0, k = 4

Explanation:

Given equation:

`y=x^2+kx+3k+4x+4`

Rearrange into standard form
`ax^2+bx+c`:

`\implies y=x^2+kx+4x+3k+4`

`\implies y=x^2+(k+4)x+(3k+4)`

Therefore:

`a=1, \quad b=(k+4), \quad c=(3k+4)`

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

`b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0`

`\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}`

`\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}`

`\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}`

Therefore, set the discriminant to zero and solve for k:

`\implies (k+4)^2-4(1)(3k+4)=0`

`\implies k^2+8k+16-12k-16=0`

`\implies k^2-4k=0`

`\implies k(k-4)=0`

`\implies k=0, k=4`

So the vertex lies in the x-axis when k = 0 or k = 4