Question

If the vertex of the parabola $y=x^2-16x+k$ lies on x−axis, then k=

Answer 1

Equation of parabola : $y=x^2-16x+k$
$\Rightarrow y=(x-8{)}^2-64+k$
$\Rightarrow y+64-k=(x-8{)}^2$
$\Rightarrow (x-8{)}^2=4\times \frac{1}{4}(y+64-k)\dots (1)$
For coordinate of vertex,
$\Rightarrow x-8=0\text{and}y+64-k=0$
$\Rightarrow x=8\text{and}y=k-64$
$\therefore \text{Vertex, V=(8, k−64)}$
Since vertex lies on the x− axis,
$\Rightarrow k-64=0$
$\Rightarrow k=64$
Hence, if the vertex of the parabola $y=x^2-16x+k$ lie on x−axis, then k = 64.