h(t) = -16t^2 + 42t + 5
find the vertex of this parabola using the method you learned
your domain is obviously t ≥ 0 to the value of t when h(t) = 0
since your h(t) is clearly above the ground, h(t) ≥ 0 up to the max of h(t), obtained at the vertex.
A baseball is thrown in the air at 5 ft. with an initial velocity of 42 ft per second. let t represent time and seconds. after the baseball is thrown let h represent height of the baseball. quadratic function h(t)= -16t squared + 42 t + 5
Find domain and range
3 answers
I got 32.5 for vertex...the rest doesn't make since to me
Here is a picture of what you have:
https://www.wolframalpha.com/input/?i=h%28t%29+%3D+-16t%5E2+%2B+42t+%2B+5
The t of your vertex is 42/32 = 21/16
that is the time it takes to reach your maximum
sub that back into the original equation, h(21/16) = 521/16 = appr 32.56
so your vertex is (21/16, 521/16)
You had the height of the vertex correct,
so for the range: 0 ≤ h(t) ≤ 32.56
to find the right side of your domain:
-16t^2 + 42 + 5 = 0
so your domain would be all values of t ≥0 up to the positive answer of your equation.
https://www.wolframalpha.com/input/?i=h%28t%29+%3D+-16t%5E2+%2B+42t+%2B+5
The t of your vertex is 42/32 = 21/16
that is the time it takes to reach your maximum
sub that back into the original equation, h(21/16) = 521/16 = appr 32.56
so your vertex is (21/16, 521/16)
You had the height of the vertex correct,
so for the range: 0 ≤ h(t) ≤ 32.56
to find the right side of your domain:
-16t^2 + 42 + 5 = 0
so your domain would be all values of t ≥0 up to the positive answer of your equation.
1 answer