Check the pictures below.
if we knew the roots/solutions of the equation, we can set h(s) = 0 and solve for "s" to find out how many seconds is it when the height is 0.
if you notice in the first picture, when f(x) = 0, is when the parabola hits a root/solution or the ground, for David he'll be hitting the water surface, and the equation that has both of those roots/solutions conspicuous is
h(s) = -4.9(s - 2)(s + 1).
1) When the denominator equals zero that is a critical point
=> x - 2 = 0 => x = 2.
So x = 2 is a critical point
2) Simplify the numerator to find an expresion of the king p(x) ≥ 0 or p(x) ≤ 0. Where p(x) equals zero you have other(s) critical point(s)
Multiply both terms:
[2x + 5] / [ x - 2] = [x - 1] / [x - 2]
for x ≠ 2 => 2x + 5 = x - 1
=> 2x - x = - 1 - 5
=> x = - 6
Then, the two critical points are x = 2 and x = - 6.
Answer: option B.
I thought it was -4 I calculated it and that’s what I got?¿
Sure.
Can you find a number that goes into both terms, 5k and 35 ?
How about 5 ?
(5k - 35) = 5 times (k - 7) .
