What are the intercepts? the domain and range? Are there any turning points or vertex? What are the coordinates for those points?

Question

for the function h(t)=-16t²

Answer 1

h(t)=-16t²

Question (1): What are the intercepts?

Answer: The x-intercept is a point wherein the y coordinate is zero and the y-intercept is point wherein the x-coordinate is zero.

if t=0 the h(0) =0

and if h(0)=0 then t=0

In this case the x-intercept and the y-intercept are the same and they are both equal to (0,0).

Question (2): The domain and range?

Answer: The graph of this quadratic equation opens down because the leading coefficient is negative (-16). the equation is in vertex form h(t) = a(t-h)² +k, where (h,k) is the vertex.

h(t) =-16(t-0)² + 0

Therefore the vertex (h,k) = (0,0), which is also the turning point.

Question (3): What are the coordinates of the turning points?

In this case there is only one turning point because it is parabola.

Turning point = (0,0)